Inter 2nd Year

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Ellipse Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Ellipse Important Questions

Question 1.
Find the value of k if 4x + y + k = 0 is a tangent to the ellipse x² + 3y² = 3. [AP Mar. 16, 15]
Solution:
Equation of the ellipse is
x² + 3y² = 3
\(\frac{x^{2}}{3}\) + \(\frac{y^{2}}{1}\) = 1
a² =, 3, b² = 1
Equation of the line is 4x + y + k = 0
y = -4x – k .
m = -4c = -k.
Condition for tangency is c² = a²m² + b²
(-k)² = 3 (4)² + 1 ,
k² = 48 + 1 = 49
k = ±7.

Question 2.
Find the equation of tangents to the ellipse 2x² + y² = 8 which are
i) Parallel to x – 2y – 4 = 0 [May. 05, Mar. 06] [T.S. Mar. 17]
Solution:
Slope will be : \(\frac{1}{2}\)
Equation of tangent y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = \(\frac{1}{2}\)x ± \(\sqrt{a^{2}\left(\frac{1}{2}\right)^{2}+b^{2}}\)
\(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{8}\) = 1
y = \(\frac{1}{2}\)x ± \(\sqrt{4 \times \frac{1}{4}+8}\)
y = \(\frac{1}{2}\)x ± 3
2y – x ± 6 = 0 required equation of tangents.
x – 2y ± 6 = 0.

Question 3.
Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is \(\frac{1}{2}\). [T.S. Mar. 15]
Solution:
Latus rectum = \(\frac{15}{2}\)
distance between foci = 2
\(\frac{2 b^{2}}{a}\) = \(\frac{15}{2}\) ; 2ae = 2
ae = 1
⇒ b² = a² – a² e²
⇒ b² = a² – 1
⇒ \(\frac{15}{4}\)a = a² – 1 .
⇒ 4a² – 15a – 4 = 0
a = 4 or a = –\(\frac{1}{4}\)
b² = a² – 1
= 16 – 1
Equation of the ellipse is \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{15}\) = 1

Question 4.
Find the equation of the ellipse in the standard form such that distance between foci is 8 and distance between directrices is 32. [Mar. 06, May. 07]
Solution:
Distance between foci = 8.
Distance between directrices = 32
2ae = 8
ae = 4
\(\frac{2 a}{e}\) = 32
\(\frac{a}{e}\) = 16
(ae) (\(\frac{a}{e}\)) = 64
a² = 64
b² = a² – a² e²
= 64 – 16 = 48
Equation of the ellipse is
∴ \(\frac{x^{2}}{64}\) + \(\frac{y^{2}}{48}\) = 1

Question 5.
Find the condition for the line x cos α + y sin α = p to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. [Mar. 14]
Solution:
Equation of the ellipse is

Question 6.
Find the equation of the ellipse with focus at (1, -1) e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0. [Mar. 05] [T.S. Mar. 19]
Solution:
P(x₁, y₁) is any point on the ellipse. Equation of the directrix is
x + y + 2 = 0
Draw PM perpendicular to ZM, Join SP
By Definition of ellipse SP = e. PM
SP² = e² . PM²
(x₁ – 1)² + (y₁ + 1)² (\(\frac{2}{3}\))²[latex]\frac{x_{1}+y_{1}+2}{\sqrt{1+1}}[/latex]²
(x₁ – 1)² + (y₁+ 1)² = \(\frac{4}{9} \frac{\left(x_{1}+y_{1}+2\right)^{2}}{2}\)
9[(x₁ – 1)² + (y₁ + 1)²] = 2 (x₁ + y₁ + 2]²
9[x₁² – 2x₁ + 1 + y₁² + 2y₁ + 1] = 2[x₁² + y₁² + 4 + 2x₁y₁ + 4x₁ + 4y₁]
9x₁² + 9y₁² – 18x₁ + 18y₁ + 18 = 2x₁² + 2y₁² + 4x₁y₁ + 8x₁ + 8y₁ + 8
7x₁² – 4x₁y₁ + 7y₁² – 26x₁ + 10y₁ + 10 = 0
focus of P (x₁, y₁) is 7x² – 4xy + 7y² – 26x + 10y + 10 = 0
This is the equation of the required Ellipse.

Question 7.
L Find tle length of major axis, minor axis, latus rectum, eccentricity, co-ordinates of centre, foci and the equations of directrices of the following ellipse. [TS Mar. 16; Mar. 14]
i) 9x² + 16y² = 144
ii) 4x² + y² – 8x + 2y + 1 = 0
iii) x² + 2y² – 4x + 12y + 14 = 0 [Mar. 11, May 07]
Solution:
Given equation is 9x² + 16y² = 144
\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1
∴ a = 4, b = 3
Length of major axis = 2a = 2 . 4 = 8
Length of minor axis = 2b = 2 . 3 = 6
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2.9}{4}\) = \(\frac{9}{2}\)
Eccentricity = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\)
Centre is C (0, 0)
Foci are (± ae, 0) = ( ± \(\sqrt{7}\), 0)
Equations of the directrices are x = ± \(\frac{a}{e}\)
x = ± 4 . \(\frac{4}{\sqrt{7}}\) = ± \(\frac{16}{\sqrt{7}}\)
\(\sqrt{7}\)x = ± 16

(ii) Given equation is 4x² + y² – 8x + 2y + 1 = 0
4(x² – 2x) + (y² + 2y) = – 1
4(x – 1)² + (y + 1)² = 4 + 1 – 1 = 4
\(\frac{(x-1)^{2}}{1}\) + \(\frac{(y+1)^{2}}{4}\) = 1
Hence a < b ⇒ y – axis is major axis
a = 1, b = 2
Length of major axis = 2b = 4
Length of minor axis = 2a = 2
Length of lattis rectum = \(\frac{2 a^{2}}{b}\) = \(\frac{2}{2}\) = 1
Eccentricity = \(\sqrt{\frac{b^{2}-a^{2}}{b^{2}}}=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}\)
Centre is C (-1, 1)
be = 2 . \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\)
Foci are (-1, 1 ± \(\sqrt{3}\))
Equations of the directrices are y + 1 = ± \(\frac{b}{e}\)
= ± \(\frac{4}{\sqrt{3}}\)
\(\sqrt{3}\) y + \(\sqrt{3}\) = ± 4
\(\sqrt{3}\) y + \(\sqrt{3}\) ± 4 = 0

iii) Given equation is x² + 2y² – 4x + 12y + 14 = 0
x² – 4x + 2 (y² + 6y) = 14
⇒ (x² – 4x + 4) + 2(y² + 6y + 9) = 4 + 18 – 14
⇒ (x – 2)² + 2(y + 3)² = 8
⇒ \(\frac{(x-2)^{2}}{8}\) + \(\frac{(y+3)^{2}}{4}\) = 1
⇒ \(\frac{(x-2)^{2}}{(2 \sqrt{2})^{2}}\) + \(\frac{(y+3)^{2}}{2^{2}}\) = 1
a = 2\(\sqrt{2}\), b = 2, h = 2, k = -3
Length of major axis = 2a = 2(2 \(\sqrt{2}\)) = 4 \(\sqrt{2}\)
Length of minor axis = 2b = 2(2) = 4
Length of latus rectum
= \(\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}\) = \(\frac{2(4)}{2 \sqrt{2}}\) = 2\(\sqrt{2}\)
Eccentricity = \(\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{4}{8}}=\frac{1}{\sqrt{2}}\)
Centre = (h, k) = (2,-3)
Foci = (h ± ae, k) = (2 ± 2, -3)
= (4, -3), (0, -3)
Equations of the directrices are x – h = ± \(\frac{a}{e}\)
x – 2 = \(\frac{2 \sqrt{2}}{\left(\frac{1}{\sqrt{2}}\right)}\)
x – 2 = ± 4
i.e., x = 6, x = -2.

Question 8.
Find the equations of tangent and normal to the ellipse 2x² + 3y² = 11 at the point whose ordinate is 1. [T.S. Mar. 16]
Solution:
Equation of the ellipse is 2x² + 3y² = 11
Given y = 1
2x² + 3 = 11
⇒ 2x² = 8
x² = 4
x = ±2
Points on the ellipse are P (2, 1) and Q(-2, 1)
Case i) P (2, 1)
Equation of the tangent is 2x . 2 + 3y . 1 = 11
4x + 3y = 11
The normal is perpendicular to the tangent Equation of the normal at P can be taken as
3x – 4y = k
The normal passes through p (2, 1)
6 – 4 = k ⇒ k = 2
Equation of the normal at P is 3x – 4y = 2.
Case ii) Q (-2, 1)
Equation of the tangent at Q is
2x(-2) + 3y . 1 = 11
-4x + 3y =1 1
4x – 3y + 11 = 0
Equation of the normal can be taken as
3x + 4y = k
The normal passes through Q (-2, 1)
-6 + 4 = k ⇒ k = -2
Equation of the normal at Q is 3x + 4y = -2
or 3x + 4y + 2 = 0.

Question 9.
Find the eccentricity, co-ordinates of foci. Length of latus rectum and equations of directrices of the following ellipses.
i) 9x² + 16y² – 36x + 32y – 92 = 0,
ii) 3x² + y² – 6x – 2y – 5 = 0 [T.S. Mar. 15]
Solution:
i) Given ellipse is
9x² + 16y² – 36x + 32y – 92 = 0
9(x² – 4x + 4) + 16 (y² + 2y + 1)
= 92 + 36 + 16
9(x – 2)² + 16(y + 1)² = 144
comparing with \(\frac{(x-2)^{2}}{16}\) + \(\frac{(y+1)^{2}}{9}\) = 1,
we get
a² = 16, b² = 9 ⇒ a = 4, b = 3.

Equations of the directrices are x = h ± \(\frac{\mathrm{a}}{\mathrm{e}}\)
x = 2 ± \(\frac{4 \times 4}{\sqrt{7}}\)
\(\sqrt{7x}\) = 2\(\sqrt{7}\) ± 16

ii) 3x² + y² – 6x – 2y – 5 = 0
Solution:
3(x² – 2x) + (y² – 2y) = 5
⇒ 3(x² – 2x + 1) + (y² – 2y + 1) = 9
⇒ 3(x – 1)² + (y – 1)² = 9
comparing with ⇒ \(\frac{(x-1)^{2}}{3}\) + \(\frac{(y-1)^{2}}{9}\) = 1,
we get
a < b ⇒ Y – axis is the major axis
a² = 3, b² = 9
a = \(\sqrt{3}\), b = 3, h = 1, k = 1
Length of major axis = 2b = 2(3) = 6
Length of minor axis = 2a = 2\(\sqrt{3}\)

Question 10.
Find the equation of the tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant. [A.P. Mar. 15, Mar. 07]
Solution:
Given ellipse is 9x² + 16y² = 144

Equation of the normal at P is

Question 11.
Show that the points of intersection of the perpendicular tangents to an ellipse lie on a circle. [A.P. Mar. 16]
Solution:
Let the equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
(a > b). Any tangent to it in the slope-intercept form is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) …………………. (1)
Let the perpendicular tangents intersect at P(x₁, y₁).
∴ P lies on (1) for some real m, i.e.,
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
∴ (y₁ – mx₁)² = a²m² + b².
or .
(x₁² – a²) m² – 2x₁y₁m + (y₁² – b²) = 0 being a
quadratic equation in ‘m’ has two roots say m₁ and m₂ then m₁, m₂ are the slopes of tangents from P to the ellipse
∴ m₁m₂ = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
∴ -1 = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
[∵ The tangents are perpendicular to each
other so that m₁m₂ = -1]
i.e., x₁² + y₁² = a² + b²,
If, however, one of the perpendicular tangents is vertical, then such pair of perpendicular tangents intersect at one of the points (± a, ± b) and any of these points satisfies x² + y² = a² + b².
∴ The point of intersection of perpendicular tangents to the ellipse S = O lies on the circle x² + y² = a² + b².

Question 12.
Find the eccentricity, co-ordinates of foci, Length of latus reEtum and equations of directrices of the following ellipses.
i) 9x² + 16y² – 36x + 32y – 92 = 0
ii) 3x² + y² – 6x – 2y – 5 = 0 [T.S. Mar. 15]
Solution:
Given ellipse is :
9x² + 16y² – 36x + 32y – 92 = 0
9(x² -4x + 4) + 16 (y² + 2y + 1)
= 92 + 36 + 16
9 (x – 2)² + 16 (y + 1)2² = 144
comparing with \(\frac{(x-2)^{2}}{16}\) + \(\frac{(y+1)^{2}}{9}\) = 1,
we get
a² = 16, b² = 9 ⇒ a = 4, b = 3.
e = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\)
Foci (h ± ae, k) = (2 ± 4 . \(\frac{\sqrt{7}}{4}\), -1)
= (2 ± \(\sqrt{7}\), -1)
Length of the latus rectum
= \(\frac{2 \cdot b^{2}}{a}=\frac{2.9}{4}=\frac{9}{2}\)
Equations of the directrices are x = h ± \(\frac{a}{e}\)
x = 2 ± \(\frac{4 \times 4}{\sqrt{7}}\)
\(\sqrt{7x}\) = 2\(\sqrt{7}\) ± 16

ii) 3x² + y² – 6x – 2y – 5 = 0
Solution:
3(x² – 2x) + (y² – 2y) = 5
⇒ 3(x² – 2x + 1) + (y² – 2y + 1) = 9
⇒ 3(x – 1)² + (y – 1)² = 9 .
comparing with ⇒ \(\frac{(x-1)^{2}}{3}\) + \(\frac{(y-1)^{2}}{9}\) = 1
we get
a < b ⇒ Y – axis is the major axis
a² = 3, b² = 9 .
a = \(\sqrt{3}\), b = 3, h = 1, k = 1
Length of major axis = 2b = 2(3) = 6
Length of minor axis = 2a = 2\(\sqrt{3}\)
Length of latus rectum = \(\frac{2 a^{2}}{b}=\frac{2.3}{3}\) = 2
Eccentricity = \(\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{3}{9}}=\sqrt{\frac{2}{3}}\)
Centre = (h, k) = (1, 1)
Focus = (h, k ± be) = (1, 1 ± 3 \(\left.\sqrt{\frac{2}{3}}\right)\))
= (1, 1 ± \(\sqrt{6}\))
Equation of directrices are y – k = ± \(\frac{b}{e}\)
y – 1 = ± \(\frac{3 \sqrt{3}}{\sqrt{2}}\)
y = 1 ± \(\frac{3 \sqrt{3}}{\sqrt{2}}\)

Question 13.
Find the equation of the ellipse referred to its major and minor axes as the co-ordinate axes x, y respectively with latus rectum of length 4 and the distance between foci 4\(\sqrt{2}\).
Solution:
Let the equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
(a > b)
Length of the latus rectum = \(\frac{2 b^{2}}{a}\) = 4
⇒ b² = 2a.
Foci are S (ae, 0), S’ (-ae, 0)
Distance between the foci = 2ae = 4\(\sqrt{2}\)
ae = 2\(\sqrt{2}\)
b² = a² (1 – e²) = a² – (ae)²
2a = a² – 8 ⇒ a² – 2a – 8 = 0
(a – 4) (a + 2) = 0
a = 4 or – 2
a > 0 ⇒ a = 4
b² = 2a = 2 . 4 = 8
Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{8}\) = 1
x² + 2y² = 16.

Question 14.
If the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse.
Solution:
Let \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a> b) be the ellipse in its standard form.
Length of latus rectum = \(\frac{1}{2}\) (minor axis)
2 \(\frac{b^{2}}{a}\) = \(\frac{1}{2}\) (2b)
2 \(\frac{b^{2}}{a}\) = b
a = 2b
a² =4 b², ⇒ a² = 4a² (1 – e²)
∴ 1 – e² = \(\frac{1}{4}\) ⇒ e² = \(\frac{3}{4}\)
⇒ e = \(\frac{\sqrt{3}}{2}\).

Question 15.
If θ₁, θ₂ are the eccentric angles of the extremeties of a focal chord (other that the vertices) of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) and e Its eccentricity. Then show that
i) e cos \(\frac{\left(\theta_{1}+\theta_{2}\right)}{2}\) = cos \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\)
ii) \(\frac{e+1}{e-1}\) = cot \(\left(\frac{\theta_{1}}{2}\right)\) . cot \(\left(\frac{\theta_{2}}{2}\right)\)
Solution:
Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1,
(a > b)

sin θ₁ . cos θ₂ – e sin θ₁ = cos θ₁ sin θ₂ – e . sin θ₂
sin θ₁ . cos θ₂ – cos θ₁ sin θ₂ = e sin θ₁ – e sin θ₂
sin (θ₁ – θ₂) = e (sin θ₁ – sin θ₂)
2 sin \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\) . cos \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\)
= e [2 cos \(\frac{\theta_{1}+\theta_{2}}{2}\) . sin \(\frac{\theta_{1}-\theta_{2}}{2}\)]

Question 16.
C is the centre1 AA’ and BB’ are major and minor axis of the ellipse.
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. If PN is the ordinate of a point P on the ellipse then show that
\(\frac{(\mathrm{PN})^{2}}{\left(\mathrm{~A}^{\prime} \mathrm{N}\right)(\mathrm{AN})}\) = \(\frac{(\mathrm{BC})^{2}}{(\mathrm{CA})^{2}}\)
Solution:

Equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
P(a cos θ, b sin θ) any point on the ellipse.
PN = b sin θ; AN = a – a cos θ,
AN = a + a cos θ; BC = b, CA = a
(A’N). (AN) = (a + a cos θ) (a – a cos θ)
= a² – a²cos²θ
= a² (1 – cos θ)
= a² sin²θ
\(\frac{(\mathrm{PN})^{2}}{\left(\mathrm{~A}^{\prime} N\right)(\mathrm{AN})}=\frac{\mathrm{b}^{2} \sin ^{2} \theta}{\mathrm{a}^{2} \sin ^{2} \theta}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\)
\(\frac{B C^{2}}{(C A)^{2}}=\frac{b^{2}}{a^{2}} \Rightarrow \frac{P^{2}}{\left(A^{1} N\right)(A N)}=\frac{(B C)^{2}}{(C A)^{2}}\)

Question 17.
S and Tare the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the eccentricity of the ellipse.
Solution:

Let \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) be an ellipse whose foci are S and T, B is an end of the minor axis such that STB is equilateral, then SB = ST = TB. We have S(ae, 0).
T = (-ae, 0) and B(0, b)
Consider SB = ST ⇒ (SB)² = (ST)²
⇒ (ae)² + b² = 4a²e²
∴ a²e² + a² (1 – e)² = 4a²e²
[∵ b² = a² (1 – e²)]
e² = \(\frac{1}{4}\)
∴ Eccentricity of the ellipse is \(\frac{1}{2}\).

Question 18.
Show that among the points on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b), (-a, 0) is the farthest point and (a, 0) is the nearest point from the focus (ae, 0).
Solution:
Let P = (x, y) be any point on the ellipse so that – a ≤ x ≤ a and S = (ae, 0) be the focus.
Since (x, y) is on the ellipse,
y² = \(\frac{b^{2}}{a^{2}}\) (a² – x²)
= (1 – e²)(a² – x²) ………….. (1) [∵ b² = a²(1 – e²)]
Then we know that
sp² = (x – ae)² + y²
= (x – ae)² + (1 – e²)(a² – x²)
= -2xae + a² + e²x²
= [a – ex]²
∴ SP = [a – ex]
we have – a ≤ x ≤ a
⇒ -ae ≤ xe ≤ ae
⇒ -ae – a ≤ xe – a ≤ ae – a …………………. (2)
∴ ex – a < 0
∴ SP = a – ex …………………… (3)
From (2) and (3)
ae + a ≥ SP ≥ a – ae
⇒ a – ae ≤ SP ≤ ae + a
∴ Max SP = ae + a when P = (-a, 0)
and Min SP = a – ae when P = (a, 0)
Hence the nearest point is (a, 0) and the farthest one is (-a, 0).

Question 19.
The orbit of the Earth is an ellipse with eccentricity \(\frac{1}{60}\) with the Sun at one of its foci, the major axis being approximately 186 × 10⁶ miles in length. Find the shortest and longest distance of the Earth from the Sun.
Solution:
We take the orbit of the Earth to be
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b).
Since the major axis is 186 × 10⁶ miles,
2a = 186 × 10⁶ miles
∴ a = 93 × 10⁶ miles
If e be the eccentricity of the orbit, e = \(\frac{1}{60}\)
We know, the longest and shortest distances of the Earth from the Sun are respectively
a + ae and a – ae (problem 7)
Here, the longest distance
= 93 × 10⁶ × (1 + \(\frac{1}{60}\))
= 9455 × 10⁴ miles.
and the shortest distance
= 93 × 10⁶ × (1 – \(\frac{1}{60}\)) miles
= 9145 × 10⁴ miles.

Question 20.
Find the equation of the tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant. A.P. [Mar. 15, Mar. 07]
Solution:
Given ellipse is 9x² + 16y² = 144

Equation of the normal at P is

Question 21.
If a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) meets its major.axis and minor axis atM and N respectively, then prove that \(\frac{a^{2}}{(C M)^{2}}\) + \(\frac{b^{2}}{(C N)^{2}}\) = 1. Where C is the centre of the ellipse.
Solution:
Let P(θ) (a cos θ, b sin θ) is any point on the ellipse then Equation of the tangent at P (θ) is

a² . \(\frac{\cos ^{2} \theta}{a^{2}}\) + b² . \(\frac{\sin ^{2} \theta}{b^{2}}\)
= cos² θ + sin² θ = 1.

Question 22.
Find the condition for the line
i) lx + my + n = 0 to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
ii) lx+ my n = 0 to be a normal to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.
Solution:
i) Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
Equation .of the tangent at P(θ) is
\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 ……………………… (1)
Equation of the given line is
lx + my = -n …………………. (2)
(1), (2) represent the same line. Comparing the co-efficients

Comparing (1) and (2)

Question 23.
If the normal at one end of a latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 passes through one end of the minor axis, then show that e⁴ + e² = 1 [e is the eccentricity of the ellipse]
Solution:
Let L be the one end of the latus rectum of
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. Then the coordinates of
L = (ae, \(\frac{b^{2}}{a}\))
Hence equation of the normal at L is

is a line passes through the one end
B’ = (0, -b)
or minor axis of \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 as shown in figure.
\(\frac{\mathrm{a}(0)}{\mathrm{e}}\) – a(-b) = a² – b²
ab = a2² – a² (1 – e²)
ab = a²e² ⇒ e² = \(\frac{b}{a}\) ⇒ e⁴ = \(\frac{b^{2}}{a^{2}}\)
= \(\frac{a^{2}\left(1-e^{2}\right)}{a^{2}}\) = 1 – e² ⇒ e⁴ + e² = 1.

Question 24.
If PN is the ordinate of a point P on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 and the tangent at P meets the X – axis at T then show that (CN) (CT) = a² where C is the centre of the ellipse.
Solution:
Let P(θ) = (acosθ, bsinθ) be a point on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. Then the equation of the tangent at P(θ) is \(\frac{x \cos \theta}{a}\) + \(\frac{y \sin \theta}{b}\) = 1 or \(\frac{x}{\frac{a}{\cos \theta}}+\frac{y}{\frac{b}{\sin \theta}}=1\) meets the X – axis at T

x – intercept (CT) = \(\frac{a}{\cos \theta}\) and the ordinate of P is PN = bsinθ
then its absicca CN = a cos θ. (see Fig.)
∴ (CN) . (CT) =, (a cos θ) (\(\frac{a}{\cos \theta}\)) = a².

Question 25.
Show that the points of intersection of the perpendicular tangets to an ellipse lie on a circle. [A.P. Mar. 16]
Solution:
Let the equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b). Any tangent to it in the slope intercept
form is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) …………………. (1)
Let the perpendicular tangents intersed at
P(x₁, y₁).
∴ p lies on (1) for some real m, i.e.,
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
∴ (y₁ – mx₁)² = a²m² + b².
or
(x₁² – a²) m² – 2x₁y₁m + (y₁² – b²) = 0 being a quadratic equation in ‘m’, has two roots say m₁ and m₂ then m₁, m₂ are the slopes of tangents from P to the ellipse
∴ m₁m₂ = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
∴ – 1 = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
[∵ The tangents are perpendicular to each other so that m₁ m₂ = -1]
i.e., \(x_{1}^{2}\) + \(y_{1}^{2}\) = a² + b².
If, however, one of the perpendicular tangents is vertical, then such pair of perpendicular tangents intersect at one of the points (a, ± b) and any of these points satisfies x² + y² = a² + b².
∴ The point of intersection of perpendicular tangents to the ellipse S = 0 lies on the circle x² + y² = a²+ b².

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Complex Numbers Important Questions

Question 1.
If z₁ = -1, z₂ = i then find Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) (AP Mar. 17) (TS Mar.’ 16; May ‘11)
Solution:
Z₁ = -1 = cos π + i sin π
⇒ Arg z₁ = π
z₂ = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
⇒ Arg z₂ = \(\frac{\pi}{2}\)
⇒ Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) = Arg z₁ – Arg z₂ = π – \(\frac{\pi}{2}\).
= \(\frac{\pi}{2}\)

Question 2.
If z = 2 – 3i, show that z² – 4z + 13 = 0. (Mar. ‘08)
Solution:
∴ z = 2 – 3i
⇒ z – 2 = -3i
⇒ (z – 2)² = (-3i)²
⇒ z² – 4z + 4 = 9i²
⇒ z² – 4z + 4 = 9(-1)
⇒ z² – 4z + 13 = 0

Question 3.
Find the multiplicative inverse of 7 + 24i. (TS Mar. 16)
Solution:
Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of (x + iy) is \(\frac{x-i y}{x^{2}+y^{2}}\)
Hence the multiplicative inverse of 7 + 24i is

Question 4.
Write the following complex numbers in the form A + iB. (2 – 3i) (3 + 4i)  (AP Mar. ’17)
Solution:
(2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i²
= 6 – i + 12
= 18 – i = 18 + i(-1)

Question 5.
Write the following complex numbers in the form A + iB. (1 + 2i)³ (TS Mar. ’17)
Solution:
(1 + 2i)³ = 1 + 3.i².2i + 3.1. 4i² + 8i³
= 1 + 6i – 12 – 8i
= -11 – 2i = (-11) + i(-2)

Question 6.
Write the conjugate of the following complex number \(\frac{5 i}{7+i}\) (AP Mar. ’15)
Solution:

Question 7.
Find a square root for the complex number 7 + 24i. (Mar. ‘14)
Solution:
7 + 24i

Question 8.
Find a square root for the complex number 3 + 4i  (Mar. ’13)
Solution:

Question 9.
Express the following complex numbers in modulus amplitude form. 1 – i (AP Mar. 15)
Solution:
1 – i
Let 1 – i = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1
r sin θ = -1
⇒ θ lies in IV quadrant .
Squaring and adding
r² (cos² θ + sin² θ) = 1 + 1 = 2
r² = 2 ⇒ r = \(\sqrt{2}\)
tan θ = -1
⇒ θ = -π/4

Question 10.
Express the complex numbers in modulus — amplitude form 1 + i\(\sqrt{3}\) (TS Mar. ’17)
Solution:
1 + i\(\sqrt{3}\) = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1 —– (1)
r sin θ = \(\sqrt{3}\) —– (2)
θ lies in I quadrant
Squaring and adding (1) and (2)
r² (cos² θ – sin² θ) = 1 + 3
r² = 4 ⇒ r = 2
Dividing (2) by (1)

Question 11.
If the Arg \(\overline{\mathbf{z}}_{1}\) and Arg \(z_{2}\) are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, find (Arg z₁ + Arg z₂) (AP Mar. ’16)
Solution:
Arg \(\overline{\mathbf{z}}_{1}\) = \(\frac{\pi}{5}\) ⇒ Arg z₁ = – Arg z₁ = – \(\frac{\pi}{5}\)
Arg z₂ = \(\frac{\pi}{3}\)
∴ Arg z₁ + Arg z₂ = – \(\frac{\pi}{5}\) + \(\frac{\pi}{3}\)
= \(\frac{-3 \pi+5 \pi}{15}\) = \(\frac{2 \pi}{15}\)

Question 12.
If |z – 3 + i| = 4 determine the locus of z. (May. ’14)
Solution:
Let z = x + iy
Given |z – 3 + i| = 4
|x + iy – 3 + i| = 4
⇒ (x – 3) + i(y + 1) = 4
⇒ \(\sqrt{(x-3)^{2}+(y+1)^{2}}\) = 4
⇒ (x – 3)² + (y + 1)² = 16
⇒ x² – 6x + 9 + y² + 2y + 1 = 16
⇒ x² + y² – 6x + 2y – 6 = 0
∴ The locus õf z is x² + y² – 6x + 2y – 6 = 0

Question 13.
The points P, Q denote the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If z₁z₂ + z₂z₁ = 0, show that POQ = 90°. (Mar. ‘07)
Solution:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

Question 14.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\). (TS Mar. 15)
Solution:

Question 15.
Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar. ’11)
Solution:

Since the given point lies in the second quadrant we look for a solution of tan θ = –\(\sqrt{3}\) that lies in \(\left[\frac{\pi}{2}, \pi\right]\), we find that θ = \(\frac{2 \pi}{3}\) is such a solution.

Question 16.
z = x + iy and the point P represents z in the Argand plane and \(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1, Re (a) ≠ 0, then find the locus of P. (TS Mar. ’17)
Solution:
Let z = x + iy and a = α + iβ

Locus of P is x = 0 i.e., Y – axis

Question 17.
If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), show that 4x² – 1 = 0 (AP Mar. ’16, TS Mar. ’17, ’15, ’06 )
Solution:

Equating real parts on both sides, we have
x = \(\frac{1}{2}\)
2x = 1
⇒ 4x² = 1
4x² – 1 = 0

Question 18.
If (\(\sqrt{3}\) + 1)¹⁰⁰ = 2⁹⁹ (a + ib), then show that a² + b² = 4. (AP Mar. ‘16)
Solution:

Question 19.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2\(\sqrt{3}\) + 2\(\sqrt{3}\)i are the vertices of an equilateral triangle. (Mar ‘07)
Solution:
A (2, 2), B (-2, -2), C (-2\(\sqrt{3}\), 2\(\sqrt{3}\)) represents the given complex number in the Argand diagram.

Question 20.
Show that the points in the Argand plane represented by the complex numbers -2 + 7i, –\(\frac{3}{2}\), +\(\frac{1}{2}\)i, 4 – 3i, \(\frac{7}{2}\)(1 + i) are the vertices of a rhombus. (June 04) (TS Mar. ’16; AP Mar.’15 ’05; May ’05)
Solution:
A(-2, 7), B(-\(\frac{3}{2}\), \(\frac{1}{2}\)), C(4, -3), D(\(\frac{7}{2}\), \(\frac{7}{2}\)) represents the given complex numbers in the Argand diagram.


∴ AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA
AC² = (-2 – 4)² + (7 + 3)²
= 36 +100 = 136
(BD)² = (-\(\frac{3}{2}\) – \(\frac{7}{2}\))² + (\(\frac{1}{2}\) – \(\frac{7}{2}\))²
= 25 + 9 = 34
AC ≠ BD
A, B, C, D are the vertices of a Rhombus.

Question 21.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃ are collinear, if and only if there exists three real numbers p, q, r not all zero, satisfying pz₁ + qz₂ + rz₃ = 0 and p + q + r = 0. (Mar. ‘07)
Solution:
pz₁ + qz₂ + rz₃ = 0
⇔ rz₃ = -pz₁ – qz₂
⇔ z₃ = \(\frac{-p z_{1}-q z_{2}}{r}\) ∵ r ≠ 0
∵ p + q + r = 0
⇔ r = -p – q
⇔ z₃ = \(-\frac{\left(p z_{1}+q z_{2}\right)}{-(p+q)}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ divides line segment joining z₁, z₂ in the ratio q : p
⇔ z₁, z₂, z₃ are collinear

Question 22.
If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus. (Mar. ’06)
Solution:
Let z = (x + iy)

Hence a = 0 and b > 0
∴ x(x – 2) + y(y – 6) = 0
or x² + y² – 2x – 6y = 0.

Question 23.
If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then show that x² + y² = 4x – 3 (TS Mar ’17)
Solution:

Equating real and imaginary parts on both sides, we have

Question 24.
Express \(\frac{4+2 i}{1-2 i}\) + \(\frac{3+4 i}{2+3 i}\) in the form a + ib, a ∈ R, b ∈ R.
Solution:

Question 25.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\) (TS Mar ’15)
Solution:

Question 26.
Express (1 – 3)³ (1 + i) in the form of a + ib.
Solution:
(1 – i)³ (1 + j) = (1 – j)² (1 – i) (1 + j)
= (1 + i² – 2i) (1² – i²)
= (1 – 1 – 2i) (1 + 1)
= 0 – 4i = 0 + i (-4)

Question 27.
Find the multiplicative inverse of 7 + 24i.  (TS. Mar. ’16 )
Solution:
Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of

Question 28.
Determine the locus of z, z ≠ 2i, such that Re\(\left(\frac{z-4}{z-2 i}\right)\) = 0
Solution:
Let z = x + iy

Hence the locus of the given point representing the complex number is the circle with (2, 1) as centre and \(\sqrt{5}\) units as radius, excluding the point (0, 2).

Question 29.
If 4x + i (3x – y) = 3 -6i where x and y are real numbers, then find the values of x and y.
Solution:
∵ 4x + i(3x – y) = 3 – 6i
Equating real and imaginary parts, we get 4x = 3 and 3x – y = -6
4x = 3 and 3x – y = -6
⇒ x = 3/4 and 3\(\left(\frac{3}{4}\right)\) – y = -6
\(\frac{9}{4}\) + 6 = y
⇒ y = \(\frac{33}{4}\)
∴ x = \(\frac{3}{4}\) and y = \(\frac{33}{4}\)

Question 30.
If z = 2 – 3i, show that z² – 4z + 13 = 0. (Mar. ’08)
Solution:
∴ z = 2 – 3i
⇒ z – 2 = – 3i
⇒ (z – 2)² = (-3i)²
⇒ z² – 4z + 4 = 9i²
⇒ z² – 4z + 13 = 0
⇒ z² – 4z + 4 = 9

Question 31.
Find the complex conjugate of (3 + 4i) (2 – 3i).
Solution:
The given complex number is
(3 + 4i) (2 – 3i) = 6 + 8i – 9i – 12i²
= 6 – i – 12(-1) = 18 + i
Its complex conjugate is 18 + i

Question 32.
Show that z₁ = \(\frac{2+11 i}{25}\), z₂ = \(\frac{-2+i}{(1-2 i)^{2}}\), are conjugate to each other.
Solution:

Since, this complex number is the conjugate of \(\frac{2+11 i}{25}\), the two given complex numbers
are conjugate to each other.

Question 33.
Find the square root of (-5 + 12i).
Solution:
We have \(\sqrt{a+i b}\) =

In this example a = -5, b = 12

Question 34.
Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar ’11)
Solution:

Since the given point lies in the second quadrant we look for a solution of
tan θ = – \(\sqrt{3}\) that lies in [\(\frac{\pi}{2}\), π] we find that θ = \(\frac{2 \pi}{3}\) is such a solution.
∴ –\(\sqrt{7}\) + i\(\sqrt{21}\) = 2\(\sqrt{7}\) cis \(\frac{2 \pi}{3}\)
(or) 2\(\sqrt{7}\)(cos \(\frac{2 \pi}{3}\) + i sin \(\frac{2 \pi}{3}\))

Question 35.
Express -1 – i in polar form with principle value of the amplitude.
Solution:
Let -1 – i = r (cos θ + i sin θ), then
-1 = r cos θ, -1 = r sin θ, tan θ = 1 ——— (1)
∴ r² = 2
⇒ r = ±\(\sqrt{2}\)
Since θ is positive, -π < θ < π, the value θ satisfying the equation (1) is
θ = -135° = \(\frac{-3 \pi}{4}\)

Question 36.
If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus.  (Mar. ’06)
Solution:

By hypothesis, amplitude of a + ib = \(\frac{\pi}{2}\)
So \(\frac{\pi}{2}\) = tan^-1 \(\frac{b}{a}\)
Hence a = 0 and b > 0
∴ x(x – 2) + y(y – 6) = 0
or x² + y² – 2x – 6y = 0.

Question 37.
Show that the equation of any circle in the complex plane is of the form z\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + c = 0, 1(b ∈ C, c ∈ R).
Solution:
Assume the general form of the equation of a circle in cartesian co-ordinates as
x² + y² + 2gx + 2fy + c = 0, (g, f ∈ R) —— (1)
To write this equation in the complex variable form, let (x, y) = z.
Then \(\frac{z+\bar{z}}{2}\) = x, \(\frac{z-\bar{z}}{2 i}\)
= y = \(\frac{-i(z-\bar{z})}{2}\)
∴ x² + y² = |z|² = z\(\overline{\mathbf{z}}\)
Substituting these results in equation (1), we obtain
z\(\overline{\mathbf{z}}\) + g(z + \(\overline{\mathbf{z}}\)) + f(z – \(\overline{\mathbf{z}}\))(-i) + c = 0
i.e., z\(\overline{\mathbf{z}}\) + (g – if)z + (g + if)\(\overline{\mathbf{z}}\) + c = 0 ——-(2)
If (g + if) = b, then equation (2) can be written as z\(\overline{\mathbf{z}}\) + \(\overline{\mathbf{b}}\)z + b\(\overline{\mathbf{z}}\) + c = 0

Question 38.
Show that the complex numbers z satisfying z² + \((\overline{\mathbf{z}})^{2}\) = 2 constitute a hyperbola.
Solution:
Substituting z = x + iy in the given equation
z² + (\(\overline{\mathbf{z}}\))² = 2, we obtain the cartesian form of the given equation.
∴ (x + iy)² + (x – iy)² = 2
i.e., x² – y² + 2ixy + x² – y² – 2ixy = 2
i.e., x² – y² = 1.
Since, this equation denotes a hyperbola, all the complex numbers satisfying

lie on the hyperbola x² – y² = 1.

Question 39.
Show that the points in the Argand diagram represented by the complex numbers 1 + 3i, 4 – 3i, 5 – 5i are collinear.
Solution:
Let the three complex numbers be represented in the Argand plane by the points P, Q, R respectively. Then P = (1, 3), Q = (4, -3), R = (5, -5). The slope of the line segment joining P,Q is \(\frac{3+3}{1-4}\) = \(\frac{6}{-3}\) = -2.
Similarly the slope of the line segment joining Q, R is \(\frac{-3+5}{4-5}\) = \(\frac{2}{-1}\) = -2.
Since the slope of PQ is the slope of QR, the points P, Q and R are collinear.

Question 40.
Find the equation of the straight line joining the points represented by (- 4 + 3i), (2 – 3i) in the Argand plane.
Solution:
Take the given points as
A = -4 + 3i = (-4, 3)
B = 2 – 3i = (2, -3)
Then equation of the straight line \(\overleftrightarrow{\mathrm{AB}}\) is
y – 3 = \(\frac{3+3}{-4-2}\)(x + 4)
i.e., x + y + 1 = 0.

Question 41.
z = x + iy represents a point in the Argand plane, find the locus of z. Such that |z| = 2.
Solution:
|z| = 2, z = x + iy
if \(\sqrt{x^{2}+y^{2}}\) = 2
if \(\sqrt{x^{2}+y^{2}}\) = 2
if and only if x² + y² = 4
The equation x² + y² = 4 represents the circle with centre at the origin (0, 0) and radius 2 units.
∴ The locus of |z| = 2 is the circle
x² + y² = 4

Question 42.
The point P represents a complex number z in the Argand plane. If the amplitude of z is \(\frac{\pi}{4}\), determine the locus of P.
Solution:
Let z = x + iy.
By hypothesis, amplitude of z = \(\frac{\pi}{4}\)
Hence tan^-1 \(\left(\frac{y}{x}\right)\) = \(\frac{\pi}{4}\) and \(\frac{y}{x}\) = tan \(\frac{\pi}{4}\)
Hence x = y
∴ The locus of P is x = y.

Question 43.
If the point P denotes the complex number z = x + iy in the Argand plane and if \(\frac{z-i}{z-1}\) is a purely imaginary number, find the locus of P.
Solution:
We note that the quotient \(\frac{z-i}{z-1}\) is not defined if z = 1.

∴ The locus of P is the circle
x² + y² – x – y = 0
excluding the point (1, 0).

Question 44.
Describe geometrically the following subsets of C.
i) {z ∈ C| |z – 1 + i| = 1}
ii) {z ∈ C| |z + i| ≤ 3|
Solution:
i) Let S = {z ∈ C| z – 1 + i| = 1}
If we write z = (x, y), then
S = {(x, y) ∈ R²||x + iy – 1 + i| = 1}
= {x, y) ∈ R² || x + i(y – 1)| ≤ 3}
= {(x, y) ∈ R² || (x – 1)² + (y + 1)² = i}
Hence S is a circle with centre (1, -1) and radius 1 unit.

ii) Let S’ = {z ∈ C || z + i| ≤ 3}
Then S = {(x, y ∈ R² || x + iy + i| ≤ 3}
= {(x, y) ∈ R² || x² + i(y + 1) ≤ 3}
= {(x, y) ∈ R² || x² + (y + 1)² ≤ 9}
Hence S’ is the closed circular disc with centre at (0, -1) and radius 3 units.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Hyperbola Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Hyperbola Important Questions

Question 1.
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8. [T.S. Mar. 16; May 11]
Solution:
Foci are S(±5,0) ∴ ae = 5
Length of transverse axis = 2a = 8
a = 4
e = \(\frac{5}{4}\)
b² = a²(e² – 1) = 16(\(\frac{25}{16}\) – 1) = 9
Equation of the hyperbola is \(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1
9x² – 16y² = 144.

Question 2.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate-hyperbola. [AP Mar. 16] [TS Mar. 15, 13]
Solution:
If e and e₁, are the eccentricity of a hyper bola and its conjugate hyperbola, then
\(\frac{1}{\mathrm{e}^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
Given e = \(\frac{5}{4}\) = \(\frac{16}{25}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
\(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 – \(\frac{16}{25}\) = \(\frac{9}{25}\) e_1² = \(\frac{25}{9}\)
e₁ = \(\frac{5}{3}\)

Question 3.
Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the x² – 4y² = 4 hyperbola. [A.P. Mar. 16; May 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{4}\)– \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
Centre is c (0, 0)
a²e² = a² + b² = 4 + 1 = 5
ae = \(\sqrt{5}\)
Foci are (±ae, 0) = (±\(\sqrt{5}\), 0)
Eccentricity = \(\frac{\mathrm{ae}}{\mathrm{a}}\) = \(\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ± \(\frac{\mathrm{ae}}{\mathrm{a}}\)
= ± 2 . \(\frac{2}{\sqrt{5}}\)
⇒ \(\sqrt{5}\) x = ± 4
⇒ \(\sqrt{5}\) x ± 4 = 0
Length of the latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2.1}{2}\) = 1

Question 4.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and
ii) Perpendicular to the line y = x – 7. [AP Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3.
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) The tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ± 1.

Question 5.
If 3x – 4y – k = 0 is a tangent to x² – 4y² = 5, find value of k. [T.S. Mar. 17]
Solution:
Equation of the hyperbola x² – 4y² = 5
\(\frac{x^{2}}{5}\) – \(\frac{y^{2}}{\left(\frac{5}{4}\right)}\) = 1
a² = 5, b² = \(\frac{5}{4}\)
Equation of the given line is 3x — 4y + k = 0
4y = 3x + k
y = \(\frac{3}{4}\) x + \(\frac{k}{4}\)
m = \(\frac{3}{4}\), c = \(\frac{k}{4}\),
Condition for tangency is c² = a²m² – b²
\(\frac{\mathrm{k}^{2}}{16}\) = 5 . \(\frac{9}{16}\) – \(\frac{5}{4}\)
k² = 45 – 20 = 25
k = ± 5.

Question 6.
Find the equations of the tangents to the hyperbola x² – 4y² = 4 Which are
i) Parallel
ii) Perpendicular to the line x + 2y = 0. [T.S. Mar. 15, Mar. 14, 11; May 06]
Solution:
Equation of the hyperbola is
x² – 4y² = 4
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
i) The tangent is parallel to x + 2y = 0
m = –\(\frac{1}{2}\)
c² = a²m² – b² = 4 . \(\frac{1}{4}\) = 1 – 1 = 0
c = 0
Equation of the parallel tangent is
y = mx + c
= –\(\frac{1}{2}\) x
2y = -x
x + 2y = 0.

ii) The tangent is perpendicular to x + 2y = 0
Slope of the tangent = m = \(\frac{-1}{\left(-\frac{1}{2}\right)}\) = 2
c² = a²m² – b² = 4 . 4 – 1 = 153
c = ±\(\sqrt{15}\)
Equation of the perpendicular tangent is
y = 2x ± \(\sqrt{15}\)

Question 7.
Prove that the point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b². [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m²(x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₁ which are the slopes of the tangents
passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = – 1 ⇒ y₁² + b²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 8.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 [Mar. 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
∴ b² = a²(e² – 1) ⇒ e² – 1 = \(\frac{b^{2}}{a^{2}}\)
e² = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{a^{2}}\)
∴ \(\frac{1}{\mathrm{e}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\) …………… (1)
Equation of the conjugate hyperbola is

Question 9.
Find the centre eccentricity, foci, directrices and length of the latus rectum of the following hyperbolas.
i) 4x² – 9y² – 8x – 32 = 0
ii) 4 (y + 3)² – 9(x – 2)² = 1.
Solution:
i) 4x² – 9y² – 8x – 32 = 0
4(x² – 2x) – 9y² = 32
4(x² – 2x + 1) – 9y² = 36
\(\frac{(x-1)^{2}}{9}\) – \(\frac{(y)^{2}}{4}\) = 1
Centre of the hyperbola is C (1, 0)
a² = 9, b² = 4 ⇒ a = 3, b = 2
e = \(\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{9+4}{9}}=\frac{\sqrt{13}}{3}\)
Foci are (1±3. \(\frac{\sqrt{13}}{3}\), 0) = (1±\(\sqrt{13}\), 0)
Equations of differences are x = 1 ± \(\frac{3.3}{\sqrt{13}}\)
⇒ x = 1 ± \(\frac{9}{\sqrt{13}}\)
Length of the latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2.4}{3}\) = \(\frac{8}{3}\)

ii) The equation of the hyperbola is
4 (y + 3)² – 9 (x – 2)² = 1
\(\frac{y-(-3)^{2}}{1 / 4}\) = \(\frac{(x-2)^{2}}{1 / 9}\) = 1
Centre is C (2, -3)
Semi transverse axis = b = \(\frac{1}{2}\)
Semi conjugate axis = a = \(\frac{1}{3}\)

Question 10.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{e_{1}^{2}}\) = 1. [Mar. 11]
Solution:

Equation of the conjugate hyperbola is

Question 11.
i) If the line lx + my = 0 is a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that a²l² – b²m² = n².

ii) If the lx + my = 1 is a normal to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that \(\frac{a^{2}}{l^{2}}\) – \(\frac{b^{2}}{m^{2}}\) = (a² + b²)².
Solution:
i) Equation of the given tangent ¡s
lx + my + n = 0 ……………. (1)
Equation of the tangent P(θ) is
\(\frac{x}{a}\) sec θ – \(\frac{y}{b}\) tan θ – 1 = 0 …………….. (2)
Comparing (1) and (2)
\(\frac{\sec \theta}{a l}\) = \(\frac{\tan \theta}{-\mathrm{bm}}\) = \(\frac{-1}{n}\)
sec θ = –\(\frac{\mathrm{a} l}{\mathrm{n}}\), tan θ = \(\frac{\mathrm{bm}}{\mathrm{n}}\)
sec² θ – tan²θ = 1
= \(\frac{a^{2} l^{2}}{n^{2}}\) – \(\frac{b^{2} m^{2}}{n^{2}}\) = 1 ⇒ a²l² – b²m² = n².

ii) Equation of the given line is lx + my = 1 ……………. (1)
Equation of the normal at P(θ) is
\(\frac{a x}{\sec \theta}\) + \(\frac{b y}{\tan \theta}\) = a² + b² ……….. (2)
Comparing (1) and (2)

Question 12.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and ii) Perpendicular to the line y = x – 7. [A.P. Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) Th tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ±1.

Question 13.
Prove that the point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b². [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m² (x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₂ which are the slopes of the tangents passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = -1 ⇒ y₁² + b² = -x₁² + a²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 14.
A circle cuts the rectangular hyperbola xy = 1 in the points (x_r, y_r), r = 1, 2, 3, 4. Prove that x₁x₂x₃x₄ = y₁y₂y₃y₄ = 1.
Solution:
Let the circle be x² + y² = a².
Since (t, \(\frac{1}{t}\)) (t ≠ 0) lies on xy = 1, the points of intersection of the circle and the hyperbola are given by
t² + \(\frac{1}{t^{2}}\) = a²
⇒ t⁴ – a²t² + 1 = 0
⇒ t⁴ + 0 . t³ – a²t² + 0 . t + 1 = 0.
If t₁, t₂, t₃ and t₄ are the roots of the above biquadratic, then t₁t₂t₃t₄ = 1.
If (x_r, y_r) = (t_r; \(\frac{1}{t_{r}}\)), r = 1, 2, 3, 4
then x₁x₂x₃x₄ = t₁t₂t₃t₄ = 1,
and y₁y₂y₃y₄ = \(\frac{1}{t_{1} t_{2} t_{3} t_{4}}\) = 1

Question 15.
If four points be taken on a rectangular hyperbola such that the chords joining any two points is perpendicular to the chord joining the other two, and if α, β, γ and δ be the inclinations to either asymptote of the straight lines joining these points to the centre, prove that
tan α tan β tan γ tan δ = 1.
Solution:
Let the equation of the rectangular hyperbola be x² – y² = a². By rotating the X-axis and the Y-axis about the origin through an angle \(\frac{\pi}{4}\) in the clockwise sense, the equation x² – y² = a² can be transformed to the form xy = c².
Let (ct_r, \(\frac{c}{t_{r}}\)), r = 1, 2, 3, 4 (t_r ≠ 0) be four point on the curve. Let the chord joining
A = (ct₁, \(\frac{c}{t_{1}}\)), B = (ct₂, \(\frac{c}{t_{2}}\)) be perpendicular to the chord joining C = (ct₃, \(\frac{c}{t_{3}}\)) and D = (ct₄, \(\frac{c}{t_{4}}\)).
The slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{\frac{c}{t_{1}}-\frac{c}{t_{2}}}{c t_{1}-c t_{2}}=\frac{-1}{t_{1} t_{2}}\)
[No chord of the hyperbola can be vertical]
Similarly slope of \(\stackrel{\leftrightarrow}{C D}\) is –\(\frac{1}{t_{3} t_{4}}\), Since \(\stackrel{\leftrightarrow}{A B}\) ⊥ \(\stackrel{\leftrightarrow}{C D}\).
\(\left(-\frac{1}{t_{1} t_{2}}\right)\left(-\frac{1}{t_{3} t_{4}}\right)\) = -1 ⇒ t₁t₂t₃t₄ = -1 ………………… (1)
We know the coordinate axes are the asymptotes of the curve, If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the positive direction of the X-axis, then tan α, tan β, tan γ and tan δ are their respective slopes. [O, the origin is the centre, None of \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) is vertical]

If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the other asymptote the Y-axis then cot α, cot β, cot γ and cot δ are their respective inclinations so that
cot α cot β cot γ cot δ = tan α tan β tan γ tan δ = 1.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 8 Differential Equations to solve questions creatively.

Intermediate 2nd Year Maths 2B Differential Equations Formulas

→ An equation involving one dependent variable and its derivatives wIth respect to one or more independent variables is called differential equation.

→ Order of a differential equation is the maximum of the orders of the derivatives.

→ Degree of a differential equation is the degree of the highest order derivative.

→ An equation involving one dependent variable, one or more independent variables and the differential coefficients (derivatives) of dependent variable with respect to independent variables is called a differential equation.

→ Order of a Differential Equation:
The order of the highest derivative involved in an ordinary differential equation is called the order of the differential equation.

→ Degree of a Differential Equation:
The degree of the highest derivative involved in an ordinary differential equation, when the equation has been expressed in the form of a polynomial in the highest derivative by eliminating radicals and fraction powers of the derivatives is called the degree of the differential equation.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 6 Integration to solve questions creatively.

Intermediate 2nd Year Maths 2B Integration Formulas

→ Integration is the inverse process of differentiation.

→ Let A ⊆ R and let f: A → R be a function. If there is a function B on A such that F'(x) = f(x), ∀ x ∈ A, then we call B an antiderivative of for a primitive of f.
i.e., \(\frac{d}{d x}\)(sin x) = cos x, ∀ x ∈ R ax
f(x) = cos x, x ∈ R, then the function
F(x) = sin x, x ∈ R is an antiderivative or primitive of f.

→ If F is an antiderivative offon A, then for k ∈ R, we have (F + k) (x) = f(x), ∀ x ∈ A.

→ Hence F + k is also an antiderivative off.
∴ c is any real number F(x + c) = G(x) = sin x + c, ∀ x ∈ R is also an antiderivative of cos x.

→ It is denoted by ∫ (cos x) dx = sin x + c, (i.e.) ∫ f(x) dx = F(x) + c.

→ Here c is called a constant of integration,
f is called the integrand and x is called the variable of integration.

Standard Forms:

→ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c if n ≠ – 1

→ ∫\(\frac{1}{x}\) dx = log |x| + c

→ ∫ sin x dx = – cos x + c, x ∈ R

→ ∫ cos x dx – sin x + c, x ∈ R

→ ∫tan x dx = log |sec x| + c

→ ∫ cot x dx = log |sin x | + c

→ ∫sec x dx = log |sec x + tan x | + c (or) log |tan \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)| + c

→ ∫cosec x dx = log |cosec x – cot x| + c (or) log |tan\(\left(\frac{x}{2}\right)\)| + c

→ ∫sec² x dx = tan x + c, x ∈ R – \(\left|\frac{n \pi}{2}\right|\), n is odd integer

→ ∫cosec² x dx = – cot x + c → R – nπ, n ∈ Z

→ ∫sec x tan x dx = sec x + c, R – \(\left[\frac{n \pi}{2}\right]\), n is an odd integer

→ ∫cosec x cot xdx = – cosec x + c, R – [nπ], n ∈ Z

→ ∫e^x dx = e^x + c, x ∈ R

→ ∫a^x dx = \(\frac{a^{x}}{\log _{e} a}\) + c, a > 0, a ≠ 1

→ ∫\(\frac{1}{\sqrt{1+x^{2}}}\) dx = sin^-1x + C = – cos^-1 (x) + c

→ ∫\(\frac{d x}{1+x^{2}}\) dx = tan^-1x + C = – cot^-1 (x) + c

→ ∫\(\frac{d x}{|x| \sqrt{x^{2}-1}}\) dx = sec^-1x + C = – cosec^-1 (x) + c

→ ∫ sinh x dx = cosh x + c

→ ∫cosh xdx = sinh x + c

→ ∫cosec²h x dx coth x + c

→ ∫sec²h x dx = tanh x + c

→ ∫cosech x coth xdx = – cosech x + c

→ ∫sech x tanh x dx = – sech x + c

→ ∫e^ax dx = \(\frac{e^{a x}}{a}\) + c

→ ∫e^ax+b dx = \(\frac{e^{a x+b}}{a}\) + c

→ ∫sin (ax + b) dx = \frac{-\cos (a x+b)}{a}\(\) + c

→ ∫cos (ax + b) dx = \(\frac{\sin (a x+b)}{a}\) + c

→ ∫sec² (ax + b) dx = \(\frac{\tan (a x+b)}{a}\) + c

→ ∫cosec² (ax + b) dx = \(\frac{-\cot (a x+b)}{a}\) + c

→ ∫cosec(ax + b) cot(ax + b) dx = \(\frac{-{cosec}(a x+b)}{a}\) + c

→ ∫sec (ax + b) tan(ax + b) dx = \(\frac{\sec (a x+b)}{a}\) + c

→ ∫f(x).g(x) dx = f(x) ∫g(x) dx – ∫[\(\frac{d}{d x}\) f(x) . ∫ g(x) dx] dx (called as integration by parts)

→ ∫\(\frac{1}{\sqrt{1+x^{2}}}\) dx = sinh^-1 (x) + c, x ∈ R = log (x + \(\sqrt{x^{2}+1}\)) + c, x ∈ R

→ ∫\(\frac{1}{\sqrt{x^{2}-1}}\) dx = cosh^-1 (x) + c (or) log (x + \(\sqrt{x^{2}-1}\)) + c, x ∈ (1, ∞)
= – cos h^-1 (- x) + c (or) log (x + \(\sqrt{x^{2}-1}\)) + c, x ∈ (1, ∞)
= log |x + \(\sqrt{x^{2}-1}\)| + c, x ∈ R – [- 1, 1]

→ ∫e^x [f(x) + f'(x)] dx = e^x. f(x) + c

→ ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) dx = sin^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\) dx = sinh^-1\(\left(\frac{x}{a}\right)\) + c (or) log \(\frac{\left|x+\sqrt{x^{2}+a^{2}}\right|}{a}\) + c

→ ∫\(\frac{1}{\sqrt{x^{2}-a^{2}}}\) dx = cosh^-1\(\left(\frac{x}{a}\right)\) + c (or) log \(\frac{\left|x+\sqrt{x^{2}-a^{2}}\right|}{a}\) + c

→ ∫\(\frac{1}{a^{2}+x^{2}}\) dx = \(\frac{1}{a}\) tan^-1\(\left(\frac{x}{a}\right)\)+ c

→ ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) = \(\frac{1}{2a}\) log \(\left|\frac{a+x}{a-x}\right|\) + C

→ ∫\(\frac{1}{x^{2}-a^{2}}\) dx = \(\frac{1}{2a}\) log \(\left|\frac{x-a}{x+a}\right|\) + c

→ ∫\(\sqrt{a^{2}-x^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{a^{2}-x^{2}}\) + \(\frac{a^{2}}{2}\) sin^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\sqrt{x^{2}+a^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{x^{2}+a^{2}}\) + \(\frac{a^{2}}{2}\) sinh^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\sqrt{x^{2}-a^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{x^{2}-a^{2}}\) + \(\frac{a^{2}}{2}\) cosh^-1\(\left(\frac{x}{a}\right)\) + c

→ To evaluate

• \(\frac{p x+q}{a x^{2}+b x+c}\) dx
• ∫ (px + q) \(\sqrt{a x^{2}+b x+c}\) dx
• ∫\(\frac{p x+q}{\sqrt{a x^{2}+b x+c}}\) dx, where a, b, c, p, q ∈ R write
  px + q = A.\(\frac{d}{d x}\) (ax² + bx + c) + B and then integrate.

→ To evaluate ∫\(\frac{d x}{(a x+b) \sqrt{p x+q}}\) where a, b, c, p, q, ∈ R put t² = px + q

→ To evaluate ∫\(\frac{1}{a+b \cos x}\) dx (or) \(\frac{1}{a+b \sin x}\) dx
(or) \(\frac{1}{a+b \cos x+c \sin x}\) dx, put tan \(\frac{x}{2}\) = t
Then sin x = \(\frac{2 t}{1+t^{2}}\), cos x = \(\frac{1-t^{2}}{1+t^{2}}\) and dx = \(\frac{2}{1+t^{2}}\) dt

→ To evaluate ∫\(\frac{a \cos x+b \sin x+c}{d \cos x+e \sin x+f}\) dx where a, b, c, d e, f ∈ R; d ≠ 0, e ≠ 0, write a cos x + b sin x + c = A [d cos x + e sin x + f]’ + B (d cos x + e sin x + f) + ∨.
Find A, B, ∨ and then integrate.

→ If I_n = ∫xⁿ . e^ax dx then I_n = \(\frac{x^{n} \cdot e^{a x}}{a}-\frac{n}{a}\) I_{n – 1} for n ∈ N

→ If I_n = ∫ sinⁿ (x) dx then I_n = – \(\frac{\sin ^{n-1}(x) \cos x}{n}\) + \(\left(\frac{n-1}{n}\right)\) I_{n – 2} for n ∈ N, n ≥ 2

→ f I_n = ∫ cosⁿ (x) dx then I_n = – \(\frac{\cos ^{n-1}(x) \sin x}{n}\) + \(\left(\frac{n-1}{n}\right)\) I_{n – 2} for n ∈ N, n ≥ 2

→ If I_n = ∫tanⁿ (x) dx then I_n = \(\frac{\tan ^{n-1}(x)}{n-1}\) I_{n – 2} for N ∈ n, n ≥ 2

→ If I_{m, n} = ∫ sin^m (x) cosⁿ (x) dx then
If I_{m, n} = \(\frac{1}{m+n}\) cos^{n – 1} (x) sin^{m + 1} (x) + \(\left(\frac{n-1}{m+n}\right)\) I_{m, n – 2} where m, n ∈ N, n ≥ 2

→ If I_{m, n} = ∫secⁿ (x) dx then I_n = \(\frac{\sec ^{n-2}(x) \tan x}{n-1}\) + \(\left(\frac{n-2}{n-1}\right)\) I_{n – 2}

Theorem: If f(x) and g(x) are two integrable functions then
∫ f(x).g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx] dx.
Proof:
\(\frac{\mathrm{d}}{\mathrm{dx}}\) [f(x). ∫g(x)dx] = f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)[∫ g(x)dx] + ∫g(x)dx .\(\frac{\mathrm{d}}{\mathrm{dx}}\)[f(x)]
= f(x)g(x) + [∫g(x)dx]f’(x)
∴ ∫[f(x)g(x) + f’(x)∫g(x)dx] dx = f(x)∫g(x)dx
⇒ ∫f (x)g(x)dx + ∫f’(x) [∫g(x)dx] dx = f (x)∫g(x) dx
∴ ∫f(x)g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx]dx

Note 1: If u and v are two functions of x then ∫u dv = uv – ∫v du.

Note 2: If u and v are two functions of x; u’, u”, u”’ …………. denote the successive derivatives of u and v₁, v₂, v₃, v₄, v₅ … the successive integrals of v then the extension of integration by pairs is
∫uv dx = uv₁ – u’v₂ + u”v₃ – u”’v₄ + ………

Note 3: In integration by parts, the first function will be taken as the following order.
Inverse functions, Logarithmic functions, Algebraic functions, Trigonometric functions and Exponential functions. (To remember this a phrase ILATE).

Theorem: ∫e^ax cos bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) (a cos bx + b sin bx) + c
Proof:

Theorem: ∫e^ax sin bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) (a sin bx – b cos bx)
Proof:
Let I = ∫e^ax sin bx dx = sin bx ∫e^ax dx – ∫[d(sin bx) ∫e^ax dx] dx

Theorem: ∫ e^x [f(x) + f’(x)]dx = e^xf(x) + c
Proof:
∫e^x [f(x) + f’(x)]dx = ∫e^x f(x)dx + ∫e^x f’(x)dx
= f(x) ∫ e^xdx – ∫[d[f(x)] ∫e^xdx] dx + ∫e^x f'(x)dx
= f(x)e^x – ∫f'(x)e^xdx + ∫e^xf'(x) dx = e^xf(x) + c
Note: ∫e^-x [f(x) – f’(x)]dx = – e^-xf(x) + c

Definition: If f(x) and g(x) are two functions such that f’(x) = g(x) then f(x) is called antiderivative or primitive of g(x) with respect to x.

Note 1: If f(x) is an antiderivative of g(x) then f(x) + c is also an antiderivative of g(x) for all c ∈ R.

Definition: If F(x) is an antiderivative of f(x) then F(x) + c, c ∈ R is called indeVinite integral of f(x) with respect to x. It is denoted by ∫f(x)dx. The real number c s called constant of integration.

Note:

• The integral of a function need not exists. If a function f(x) integral then f(x) is called an integrable function.
• The process of finding the integral of a function is known as Integration.
• The integration is the reverse process of differentiation.

Corollary:
If f(x), g(x) are two integrable functions then ∫(f ± g) (x) dx = ∫f(x)dx ± ∫fg(x)dx

Corollary:
If f₁(x), f₂(x), ……, f_n(x) are integrable functions then
∫(f₁ + f₂ + …….. + f_n)(x)dx = ∫f₁(x)dx + ∫f₂(x)dx + ……. + ∫f_n(x)dx.

Corollary:
If f(x), g(x) are two integrable functions and k, l are two real numbers then ∫(kf + lg) (x)dx = k∫f(x) dx + 1∫g(x)dx.

Theorem: If f f(x)dx = g(x) and a ≠ 0 then ∫ f(ax + b)dx = \(\frac{1}{a}\)g(ax+b)+c.
Proof:
Put ax + b = t.

Theorem: It f(x) is a differentiable function then ∫\(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + c.
Proof:
Put f(x) = t ⇒ f’(x) = \(\frac{d \mathrm{t}}{\mathrm{dx}}\) ⇒ f’(x)dx = dt
∴ ∫\(\frac{f^{\prime}(x)}{f(x)}\) = ∫latex]\frac{1}{\mathrm{t}}[/latex] dt = log |t| + c = log |f(x)| + c

Theorem: ∫tan x dx = log |sec x| for x ≠ (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Proof:
∫tan x dx = ∫\(\frac{\sin x}{\cos x}\) dx = -∫\(\frac{d(\cos x)}{\cos x}\) dx
= – log |cos x| + c = log\(\frac{1}{|\cos x|}\) + c = log|sec x| + c

Theorem: ∫cot x dx = log |sin x| + c for x ≠ nπ, n ∈ Z.
Proof:
∫cot x dx = ∫\(\frac{\cos x}{\sin x}\) dx = log |sin x| + c

Theorem: ∫ sec x dx = log |sec x + tan x| + c = log |tan(π/4 + x/2) + c for x ≠ (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Proof:

Theorem: ∫csc x dx = log|csc x – cot x| + c = log |tan x/2| + c for x ≠ nπ, n ∈ Z.
Proof:
∫csc x dx = \(\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}\) dx
= \(\int \frac{\csc ^{2} x-\csc x \cot x}{\csc x-\cot x}\) dx = log |csc x – cot x| + c
= log\(\left|\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right|\) + c
= log\(\left|\frac{1-\cos x}{\sin x}\right|\) + c
= log\(\left|\frac{2 \sin ^{2} x / 2}{2 \sin x / 2 \cos x / 2}\right|\) + c
= log |tan x/2| + c

Theorem: If f(x) is differentiable function and n ≠ – 1 then ∫[f(x)]ⁿ f’(x)dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + c.
Proof:
Put f(x) = t ⇒ f’(x) dx = dt

Theorem: If ∫f(x)dx = F(x) and g(x) is a differentiable function then ∫ (fog)(x)g’(x) dx = F[g(x)] + c.
Proof:
g(x) = t ⇒ g’(x) dx = dt
∴ ∫(fog)(x)g’(x)dx = ∫f[g(x)]g’(x) dx
= ∫f(t)dt = F(t) + c = F[g(x)] + c

Theorem: ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) dx = Sin^-1\(\) + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem: ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\)dx = Sinh^-1 \(\) + c for x ∈ R.
Proof:
Put x = a sinhθ. Then dx = a cos hθ dθ
∴ ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\) dx = \(\int \frac{1}{\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}}\) a coshθ dθ
= ∫\(\frac{a \cosh \theta}{a \cosh \theta}\) = ∫dθ = θ + c = Sinh^-1\(\left(\frac{x}{a}\right)\) + c

Theorem:
∫\(\)dx = Cosh^-1\(\) + c for x ∈ (- ∞, – a) ∪ (a, ∞)
Proof:
Put x = a coshθ. Then dx = a sin hθ dθ
∴ ∫\(\frac{1}{\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}}\) dx = ∫\(\frac{1}{\sqrt{a^{2} \cosh ^{2} \theta-a^{2}}}\) a sin hθ dθ
= ∫\(\frac{a \sinh \theta}{a \sinh \theta}\) dθ = ∫ dθ = θ + c = Cosh^-1\(\left(\frac{x}{a}\right)\) + c

Theorem:
∫\(\frac{1}{a^{2}+x^{2}}\) dx = \(\frac{1}{a}\) Tan^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ R.
Proof:
Put x = a tan θ. Then dx = a sec²θ dθ

Theorem:
∫\(\frac{1}{a^{2}-x^{2}}\)dx = \(\frac{1}{2 \mathrm{a}}\) log\(\left|\frac{a+x}{a-x}\right|\) + c for x ≠ ± a
Proof:
∫\(\frac{1}{a^{2}-x^{2}}\)dx = ∫\(\left|\frac{a+x}{a-x}\right|\)dx
= \(\frac{1}{2 \mathrm{a}} \int\left(\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{a}-\mathrm{x}}\right)\)dx = \(\frac{1}{2 \mathrm{a}}\) [log |a + x| – log |a – x|] + c
= \(\frac{1}{2 \mathrm{a}}\) log\(\left|\frac{a+x}{a-x}\right|\) + c

Theorem:
∫\(\frac{1}{x^{2}-a^{2}}\) dx = \(\frac{1}{2 \mathrm{a}}\) log \(\left|\frac{x-a}{x+a}\right|\) + c for x ≠± a
Proof:
∫\(\frac{1}{x^{2}-a^{2}}\) dx = ∫\(\frac{1}{(x-a)(x+a)}\) dx
= \(\frac{1}{2 a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)\) dx = \(\frac{1}{2 \mathrm{a}}\) [log |x – a| – log |x + a|] + c
= \(\frac{1}{2 \mathrm{a}}\) log \(\left|\frac{x-a}{x+a}\right|\) + c

Theorem:
∫\(\sqrt{a^{2}-x^{2}}\)dx = \(\frac{x}{2} \sqrt{a^{2}-x^{2}}\) + \(\frac{\mathrm{a}^{2}}{2}\) sin^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem:
∫\(\sqrt{a^{2}+x^{2}}\) dx = \(\frac{x}{2} \sqrt{a^{2}+x^{2}}\) + \(\frac{\mathrm{a}^{2}}{2}\) Sinh^-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\) + c for x ∈ R.
Proof:
Put x = sinhθ. Then dx = a coshθ dθ
∴ ∫\(\sqrt{a^{2}+x^{2}}\) dx = ∫\(\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}\) a coshθ dθ
= ∫\(a \sqrt{1+\sinh ^{2} \theta}\) a coshθ dθ = a² ∫cosh² θdθ
= \(=\mathrm{a}^{2} \int \frac{1+\cosh 2 \theta}{2} \mathrm{~d} \theta=\frac{\mathrm{a}^{2}}{2}\left[\theta+\frac{1}{2} \sinh 2 \theta\right]+\mathrm{c}\)
= \(\frac{a^{2}}{2}\left[\theta+\frac{1}{2} 2 \sinh \theta \cosh \theta\right]+c\)
= \(\frac{a^{2}}{2}\left[\theta+\sinh \theta \sqrt{1+\sinh ^{2} \theta}\right]+c\)
= \(\frac{a^{2}}{2}\left[{Sinh}^{-1}\left(\frac{x}{a}\right)+\frac{x}{a} \sqrt{1+\frac{x^{2}}{a^{2}}}\right]+c\)
= \(\frac{\mathrm{a}^{2}}{2}{Sinh}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\frac{\mathrm{x}}{\mathrm{a}} \sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}+\mathrm{c}\)

Theorem:
∫\(\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}\) dx = \(\frac{x}{2} \sqrt{x^{2}-a^{2}}\) – \(\frac{a^{2}}{2}\) Cosh^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ [a, ∞)
Proof:
Put x = a coshθ. Then dx = a sinhθ dθ

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 7 Definite Integrals to solve questions creatively.

Intermediate 2nd Year Maths 2B Definite Integrals Formulas

→ If f is a function integrable over an interval [a, b] and F is a primitive offon [a, b] then
\(\int_{a}^{b}\)f(x)dx = F(b) – F(a)

→ If a < b be real numbers and y = f(x) denote a curve in the plane as shown in figure. Then the definite integral \(\int_{a}^{b}\) f(x) dx is equal to the area of the region bounded by the curve y = f(x), the ordinates x = a, x = b and the portion of X-axis.

→ \(\int_{a}^{b}\)f(x) dx = — \(\int_{b}^{a}\)f(x) dx

→ \(\int_{a}^{b}\) f(x) dx = \(\int_{a}^{c}\) f(x) dx + \(\int_{c}^{b}\)f(x) dx ; a < c < b

→ \(\int_{a}^{b}\)f[g(x)] g’ (x) dx = \(\int_{g(a)}^{g(b)}\)f(x)dx

→ \(\int_{0}^{a}\)f(x)dx = \(\int_{0}^{a}\)f(a-x)dx

→ \(\int_{0}^{2 a}\)f(x)dx =2\(\int_{0}^{a}\)f(x)dx , if f(2a-x) = f(x)
= 0, if f(2a – x) = -f(x)

→ \(\int_{-a}^{a}\)f(x)dx = 2\(\int_{0}^{a}\)f(x)dx if f is even
= 0, if f is odd

→ Let f(x) be a function defined on [a, b]. If ∫ f(x)dx = F(x), then F(b) – F(a) is called the definite integral of f(x) over [a, b]. It is denoted by \(\int_{a}^{b}\) f(x)dx. The real number ‘a’ is called the lower limit and the real number ‘b’ is called the upper limit.
This is known as fundamental theorem of integral calculus.

Geometrical Interpretation of Definite Integral:
If f(x)>0 for all x in [a, b] then \(\int_{a}^{b}\) f (x)dx is numerically equal to the area bounded by the curve y =f(x), the x-axis and the lines x = a and x = b i.e., \(\int_{a}^{b}\) f (x) dx

Properties of Definite Integrals:

• \(\int_{a}^{b}\)f (x)dx = \(\int_{a}^{b}\)f(t)dt i.e., definite integral is independent of its variable.
• \(\int_{a}^{b}\)f(x) dx = – \(\int_{b}^{a}\)f(x) dx
• \(\int_{a}^{b}\) f(x) dx = \(\int_{a}^{c}\) f(x) dx + \(\int_{c}^{b}\)f(x) dx ; a < c < b
• \(\int_{a}^{b}\)f[g(x)] g’ (x) dx = \(\int_{g(a)}^{g(b)}\)f(x)dx
• \(\int_{0}^{a}\)f(x)dx = \(\int_{0}^{a}\)f(a-x)dx
• \(\int_{0}^{2 a}\)f(x)dx =2\(\int_{0}^{a}\)f(x)dx , if f(2a-x) = f(x)
  = 0, if f(2a – x) = -f(x)
• \(\int_{-a}^{a}\)f(x)dx = 2\(\int_{0}^{a}\)f(x)dx if f is even
  = 0, if f is odd

Theorem:
If f(x) is an integrable function on [a, b] and g(x) is derivable on [a, b] then
\(\int_{a}^{b}\)(fog)(x)g'(x)dx = \(\int_{g(a)}^{g(b)}\)f(x) dx \(\int_{a}^{b}\)(fog)(x)g'(x)dx = \(\int_{g(a)}^{g(b)}\)f(x) dx

Areas Under Curves:
1. Let f be a continuous curve over [a, b]. If f(x) ≥ o in [a, b], then the area of the region bounded by y = f(x), x-axis and the lines x=a and x=b is given by \(\int_{a}^{b}\)f(x)dx.

2. Let f be a continuous curve over [a, b]. If f (x) ≤ o in [a, b], then the area of the region bounded by y = f(x), x-axis and the lines x=a and x=b is given by –\(\int_{a}^{b}\) f (x)dx.

3. Let f be a continuous curve over [a,b]. If f (x) ≥ o in [a, c] and f (x) ≤ o in [c, b] where a < c < b. Then the area of the region bounded by the curve y = f(x), the x-axis and the lines x=a and x=b is given by \(\int_{a}^{c}\)f (x )dx – \(\int_{c}^{b}\)f (x )dx

4. Let f(x) and g(x) be two continuous functions over [a, b]. Then the area of the region bounded by the curves y = f (x), y = g (x) and the lines x = a, x=b is given by |\(\int_{a}^{b}\)f(x)dx – \(\int_{a}^{b}\)g(x)dx|

5. Let f(x) and g(x) be two continuous functions over [a, b] and c ∈ (a, b). If f (x) > g (x) in (a, c) and f (x) < g (x) in (c, b) then the area of the region bounded by the curves y = f(x) and y= g(x) and the lines x=a, x=b is given by |\(\int_{a}^{c}\)(f (x) – g (x ))dx| + |\(\int_{c}^{b}\)(g(x)-f (x))dx|

6. let f(x) and g(x) be two continuous functions over [a, bi and these two curves are intersecting at X =x₁ and x = x₂ where x₁, x ₂, ∈ (a,b) then the area of the region bounded by the curves y= f(x) and y = g(x) and the lines x = x₁, x = x₂ is given by |\(\int_{x_{1}}^{x_{2}}\)(f(x) – g(x))dx|

Note: The area of the region bounded by x =g(y) where g is non negative continuous function in [c, d], the y axis and the lines y = c and y = d is given by \(\int_{c}^{d}\)g(y)dy .

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 5 Hyperbola to solve questions creatively.

Intermediate 2nd Year Maths 2B Hyperbola Formulas

Definition:
A conic with eccentricity greater than one is called a hyperbola, i.e., the locus of a point, the ratio of whose distances from a fixed point (focus) and fixed straight line (directrix) is e, where e> 1 is called hyperbola.

Equation of hyperbola in standard form:

→ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Here b² = a² (e² – 1)

→ The difference of the focal distances of any point on the hyperbola is constant.

(i.e.,) SP ~ S’P = 2a

→ A Point P(x₁, y₁) in the plane of the hyperbola S = 0 lies inside the hyperbola, if S₁₁ < 0, lies outside if S₁₁ > 0 and on the curve S₁₁ = 0

→ Ends of the latus rectum (± ae, ± b²/a) and the length of the latus rectum is 2b²/a.

→ Equation of tangent at P(x₁, y₁) to \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
\(\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}\) = 1

→ Equation of normal at P(x₁, y₁)
\(\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}\) = a² + b²

→ Parametric form is x = a sec θ, y = b tan θ.

→ Equation of tangent at ‘θ’ on the hyperbola
\(\frac{x}{a}\) sec θ – \(\frac{y}{b}\) tan θ = 1

→ Equation of normal at θ on the hyperbola
\(\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}\) = a² + b²

→ A necessary and sufficient condition for a straight line y = mx + c to be tangent to hyperbola c² = a²m² – b²
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

→ If P(x₁, y₁) is any point in the plane of the hyperbola S = 0 then the equation of the polar of P is S₁ = 0.

→ The pole of the line lx + my + n = 0; n ≠ 0 with respect to the hyperbola S = 0 is
\(\left(\frac{-a^{2} l}{n} ; \frac{b^{2} m}{n}\right)\), n ≠ 0

Equation of a Hyperbola in Standard From.
The equation of a hyperbola in the standard form is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Proof:
Let S be the focus, e be the eccentricity and L = 0 be the directrix of the hyperbola.
Let P be a point on the hyperbola. Let M, Z be the projections of P, S on the directrix L = 0 respectively. Let N be the projection of P on SZ. Since e > 1, we can divide SZ both internally and externally in the ratio e : 1.

Let A, A’ be the points of division of SZ in the ratio e : 1 internally and externally respectively. Let AA’ = 2a. Let C be the midpoint of AA’. The points A, A’ lie on the hyperbola and \(\frac{\mathrm{SA}}{\mathrm{AZ}}\) = e, \(\frac{\mathrm{SA}^{\prime}}{\mathrm{A}^{\prime} \mathrm{Z}}\) = e

∴ SA = eAZ, SA’ = eA’Z
Now SA + SA’ = eAZ + eA’Z
⇒ CS – CA + CS + CA’ = e(AZ + A’Z)
⇒ 2CS = eAA’ (∵ CA = CA’)
⇒ 2CS = e2a ^ CS = ae
Aslo SA’ – SA = eA’Z – eAZ
AA’ = e(A’Z – AZ)
⇒ 2a = e[CA’ + CZ – (CA – CZ)]
⇒ 2a = e 2CZ (∵ CA = CA’) ⇒ CZ = \(\frac{a}{e}\).

Take CS, the principal axis of the hyperbola as
x-axis and Cy perpendicular to CS as y-axis. Then S = (ae, 0).
Let P(x₁, y₁).
Now PM = NZ = CN – CZ = x₁ – \(\frac{a}{e}\).
P lies on the hyperbola ⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = e
⇒ PS = ePM ⇒ PS² = e²PM²
⇒ (x₁ – ae)² + (y₁ – 0)² = e² \(\left(x_{1}-\frac{a}{e}\right)^{2}\)
⇒ (x₁ – ae)² + y₁² = (x₁e – a)²
⇒ x₁² + a² e² – 2x₁ae + y₁² = x₁e² + a² – 2x₁ae
⇒ x²(e² -1) – y² = a²(e² -1)
⇒ \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{a^{2}\left(e^{2}-1\right)}\) = 1 ⇒ \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}\) = 1
Where b² = a²(e² – 1)
The locus of P is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
The equation of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Nature of the curve \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

Let C be the curve represented by \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1. Then
(i) (x,y) ∈ C(x, -y) ∈ C and (x, y) ∈ C ⇔ (-x,y) ∈ C.
Thus the curve is symmetric with respect to both the x-axis and the y-axis. Hence the coordinate axes are two axes of the hyperbola.

(ii) (x, y) ∈ C ⇔ (-x, -y) ∈ C.
Thus the curve is symmetric about the origin O and hence O is the midpoint of every chord of the hyperbola through O. Therefore the origin is the center of the hyperbola.

(iii) (x, y) ∈ C and y = 0 ⇒ x² = a² ⇒ x = ±a.
Thus the curve meets x-axis (Principal axis) at two points A(a, 0), A'(-a, 0). Hence hyperbola has two vertices. The axis AA’ is called transverse axis. The length of transverse axis is AA’ = 2a.

(iv) (x, y) ∈ C and x = 0 ⇒ y² = -b² ⇒ y is imaginary.
Thus the curve does not meet the y-axis. The points B(0, b), B'(0, -b) are two points on y-axis. The axis BB’ is called conjugate axis. BB’ = 2b is called the length of conjugate axis.

(v) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 ⇒ y = \(\frac{b}{a} \sqrt{x^{2}-a^{2}}\) ⇒ y has no real value for -a < x < a.
Thus the curve does not lie between x = -a and x = a.
Further x → ∞ ⇒ y ^ ± ∞ and
x → -∞ ⇒ y → ± ∞.
Thus the curve is not bounded (closed) on both the sides of the axes.

(vi) The focus of the hyperbola is S(ae, 0). The image of S with respect to the conjugate axis is S'(-ae, 0). The point S’ is called second focus of the hyperbola.

(vii) The directrix of the hyperbola is x = a/e. The image of x = a/e with respect to the conjugate axis is x = -a/e. The line x = -a/e is called second directrix of the hyperbola corresponding to the second focus S’.

Theorem:
The length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{2 b^{2}}{a}\)
Proof:
Let LL be the length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

If SL = 1, then L = (ae, 1)
L lies on the hyperbola ⇒ \(\frac{(a \mathrm{e})^{2}}{a^{2}}-\frac{1^{2}}{b^{2}}\) = 1
⇒ \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = e² – 1 ⇒ 1² = b²(e² – 1)
⇒ 1 = b² × \(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\) ⇒ 1 = \(\frac{b^{2}}{a}\) ⇒ SL = \(\frac{b^{2}}{a}\)
∴ LL’ = 2SL = \(\frac{2 b^{2}}{a}\)

Theorem:
The difference of the focal distances of any point on the hyperbola is constant f P is appoint on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 with foci S and S then |PS’- PS| = 2a
Proof:
Let e be the eccentricity and L = 0, L’ = 0 be the directrices of the hyperbola.
Let C be the centre and A, A’ be the vertices of the hyperbola.
∴ AA’ = 2a

Foci of the hyperbola are S(ae, 0), S'(-ae, 0).
Let P(x₁, y₁) be a point on the hyperbola.
Let M, M’ be the projections of P on the directrices L = 0, L’ = 0 respectively.
∴ \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, \(\frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}\) = e

Let Z, Z’ be the points of intersection of transverse axis with directrices.
∴ MM’ = ZZ’ = CZ + CZ’ = 2a/e
PS’ – PS = ePM’ – ePM = e(PM’ – PM)
= e(MM’) = e(2a/e) = 2a

Notation: We use the following notation in this chapter.
S ≡ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) – 1, S₁ = \(\frac{x_{1}}{a^{2}}-\frac{y_{1}}{b^{2}}\) – 1
S₁₁ ≡ S(x1, y1) = \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}\) – 1, S₁₂ = \(\frac{\mathrm{x}_{1} \mathrm{x}_{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}_{1} \mathrm{y}_{2}}{\mathrm{~b}^{2}}\) – 1

Note: Let F(x_1, y₁) be a point and S ≡ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) – 1 = 0 be a hyperbola. Then

• F lies on the hyperbola S = 0 ⇔ S₁₁ = 0
• F lies inside the hyperbola S = 0 ⇔ S₁₁ > 0
• F lies outside the hyperbola S = 0 ⇔ S₁₁ < 0

Theorem:
The equation of the chord joining the two points A(x₁, y₁), B(x₂, y₂) on the hyperbola S = 0 is S₁ + S₂ = S₁₂.

Theorem:
The equation of the normal to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 at P(x₁ y₁) is \(\frac{\mathrm{a}^{2} \mathrm{x}}{\mathrm{x}_{1}}+\frac{\mathrm{b}^{2} \mathrm{y}}{\mathrm{y}_{1}}\) = a² + b².

Theorem: The condition that the line y = mx + c may be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is c² = a²m² – b²

Note: The equation of the tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 may be taken as y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
The point of contact is \(\left(\frac{-a^{2} m}{c}, \frac{-b^{2}}{c}\right)\) where c = am – b

Theorem: Two tangents can be drawn to a hyperbola from an external point.
Note: If m₁, m₂ are the slopes of the tangents through P, then m₁ m₂ become the roots of (x₂² – a²)m² – 2x₁y₁m + (y₁² + b²) = 0
Hence m₁ + m₂ = \(\frac{2 \mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\)
m₁m₂ = \(\frac{\mathrm{y}_{1}^{2}+\mathrm{b}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\)

Theorem:
The point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b².
Proof:
Equation of any tangent to the hyperbola is:
y = mx ± \(\)
Suppose P(x₁, y₁) is the point of intersection of tangents.
Plies on the tangent Y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\) y₁ = mx₁ = ±\(\sqrt{a^{2} m^{2}-b^{2}}\)
= (y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ – a²m² + b² = 0
= m²(x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0

This is a quadratic in m giving the values for m say m1 and m2.
The tangents are perpendicular
⇒ m₁m₂ = -1 ⇒ \(\frac{\mathrm{y}_{1}^{2}+\mathrm{b}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\) = -1
⇒ y₁² + b² = -x² + a² ⇒ x₁² + y₁² = a² – b²
P(x₁, y₁) lies on the circle x² + y = a² – b².

Definition:
The point of intersection of perpendicular tangents to a hyperbola lies on a circle, concentric with the hyperbola. This circle is called director circle of the hyperbola.

Corollary:
The equation to the auxiliary circle of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\)= 1 is x² + y² = a².

Theorem:
The equation to the chord of contact of P(x₁, y₁) with respect to the hyperbola S = 0 is S₁ = 0.

Midpoint of a Chord:
Theorem: The equation of the chord of the hyperbola S = 0 having P(x₁, y₁) as it’s midpoint is S₁ = S₁₁.

Pair of Tangents:
Theorem: The equation to the pair of tangents to the hyperbola S = 0 from P(x₁, y₁) is S₁² = S₁₁S

Asymptotes:
Definition: The tangents of a hyperbola which touch the hyperbola at infinity are called asymptotes of the hyperbola.

Note:

• The equation to the pair of asymptotes of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 0.
• The equation to the pair of asymptotes and the hyperbola differ by a constant.
• Asymptotes of a hyperbola passes through the centre of the hyperbola.
• Asymptotes are equally inclined to the axes of the hyperbola.
• Any straight line parallel to an asymptote of a hyperbola intersects the hyperbola at only one point.

Theorem:
The angle between the asymptotes of the hyperbola S = 0 is 2tan^-1(b/a).
Proof:
The equations to the asymptotes are y = ± \(\frac{b}{a}\)x.
If θ is an angle between the asymptotes, then
tan θ = \(\frac{\frac{\mathrm{b}}{\mathrm{a}}-\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)}{1+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)}=\frac{2\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}\) = tan 2α Where tan α = \(\frac{b}{a}\)
∴ θ = 2α = 2Tan^-1\(\frac{b}{a}\)

Parametric Equations:
A point (x, y) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 represented as x = a sec θ, y = b tan θ in a single parameter θ. These equations x = a sec0, y= b tan θ are called parametric equations of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
The point (a sec θ, b tan θ) is simply denoted by θ.

Note: A point on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 can also be represented by (a cosh θ, b sinh θ). The equations x = a cosh θ, y = sinh θ are also called parametric equations of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

Theorem: The equation of the chord joining two points α and β on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is:
\(\frac{x}{a}\)cos\(\frac{\alpha-\beta}{2}\) – \(\frac{y}{b}\)sin\(\frac{\alpha+\beta}{2}\) = cos\(\frac{\alpha+\beta}{2}\)

Theorem:
The equation of the tangent at P(0) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{x}{a}\)cos\(sec θ – [latex]\frac{y}{b}\) tan θ = 1.

Theorem:
The equation of the normal at P(0) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}\) = a² + b².

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 4 Ellipse to solve questions creatively.

Intermediate 2nd Year Maths 2B Ellipse Formulas

Definition:
→ A conic with eccentricity less than one is called an ellipse i.e., the locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ which is less than 1, is called an ellipse. The fixed point and fixed straight line are called focus and directrix respectively.

Equation of Ellipse in standard form:
→ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1; a > b, b² = a² (1 – e²) ; e < 1
Foci: (ae, 0), (- ae, 0); directrices x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = \(\frac{-a}{e}\)
a = Length of semi major axis.
b = Length of semi minor axis.

Various forms of the ellipse:

→ Major axis: along X – axis
Length of major axis: 2a
Minor axis: along Y-axis
Length of minor axis: 2b
Centre: (0, 0)
Foci: S (ae, 0); S’ (- ae, 0)
Directrices: x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = – \(\frac{\mathrm{a}}{\mathrm{e}}\)
e is given as: b² = a² (1 – e²)
(or) e = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}\), a > b
a < b (or) b > a

→ Major axis: along Y – axis
Length of major axis: 2b
Minor axis: along Y-axis
Length of minor axis: 2a
Centre: (0, 0)
Foci: S (0, be); S’ (0, – be)
Directrices: y = \(\frac{b}{\mathrm{e}}\), y = – \(\frac{b}{\mathrm{e}}\)
e is given as: a² = b² (1 – e²)
(or) e = \(\sqrt{\frac{b^{2}-a^{2}}{b^{2}}} .\)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a > b
Major axis: parallel to X-axis along the line y = β
Length of major axis: 2a
Minor axis: parallel to Y-axis along the line x = α
Length of minor axis: 2b
Center: (α, β)
Foci: S (ae + α, β); S’ ( – ae + α, β)
Directrices: x = α + \(\frac{a}{e}\) ; x = α – \(\frac{a}{e}\)
e is given by b² = a² ( 1 – e²)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a < b
Major axis: parallel to Y-axis along the line x = α
Length of major axis: 2b
Minor axis: parallel to X-axis along the line y = β
Length of minor axis: 2a
Center: (α, β)
Foci: S (α, be + β); S’ (α, -be + β)
Directrices: y – β = \(\frac{b}{\mathrm{e}}\) ; y – β = – \(\frac{b}{\mathrm{e}}\)

→ A line segment joining two points on the ellipse is called a chord of the ellipse. Chord passing through foci is called a focal chord. A focal chord perpendicular to the major axis of the ellipse is cal fed latus rectum. Length of latus rectum = \(\frac{2 b^{2}}{a}\), a > b, Length of the latus rectum = \(\frac{2 a^{2}}{a}\), if b > a

→ If P (x, y) is any point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 whose foci are S and S’.
Then SP + SP’ = constant = 2a.

→ Parametric equations x = a cos θ; y = b sin θ

→ If P a point lies outside the ellipse, then S₁₁ > 0.

→ If P a point lies on the ellipse, then S₁₁ = 0.

→ If P a point lies inside the ellipse, then S₁₁ < 0.

Ellipse:
A conic is said to be an ellipse if it’s eccentricity e is less than 1.

Equation of an Ellipse in Standard Form:
The equation of an ellipse in the standard form is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1.(a < b)
Proof:

Let S be the focus, e be the eccentricity and L = 0 be the directrix of the ellipse. Let P be a point on the ellipse. Let M, Z be the projections (foot of the perpendiculars) of P, S on the directrix L = 0 respectively. Let N be the projection of P on SZ. Since e < 1, we can divide SZ both internally and externally in the ratio e: 1. Let A, A’ be the points of division of SZ in the ratio e: 1 internally and externally respectively. Let AA’ = 2a. Let C be the midpoint of AA’. The points A, A’ lie on the ellipse and
\(\frac{S A}{A Z}\) = eAZ, \(\frac{\mathrm{SA}^{\prime}}{\mathrm{A}^{\prime} \mathrm{Z}}\) = eA’Z
Now SA + SA’ = eAZ + eA’Z
⇒ AA’ = e(AZ + A’Z)
⇒ 2a = e(CZ – CA + A’C + CZ)
⇒ 2a = e.2CZ (∵ CA = A’C)
⇒ CZ = a/e

Also SA’- SA = eA’Z – eAZ
⇒ A’C + CS – (CA – CS) = e(A’Z – AZ)
⇒ 2CS = eAA’ (∵ CA = A’C)
⇒ 2CS = e2a ⇒ CS = ae
Now PM = NZ = CZ – CN = \(\frac{a}{e}\) – x₁

P lies on the ellipse:
⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = e ⇒ PS = ePM ⇒ PS² = e²PM²
⇒ (x₁ – ae)² + (y₁ – 0)² = e²\(\left(\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{x}_{1}\right)^{2}\)
⇒ (x₁ – ae)² + y₁² = (a – x₁e)²
⇒ x₁ + ae – 2x₁ae + y₁ = a + x₁e – 2x₁ae
⇒ (1 – e²)x₁² + y₁² = (1 – e²)a²
⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{a^{2}\left(1-e^{2}\right)}\) = 1 ⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}\) = 1
Where b² = a²(1 – e²) > 0
The locus of P is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
∴ The equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Nature of the Curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.

Let be the curve represented by \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1. Then

• The curve is symmetric about the coordinate axes.
• The curve is symmetric about the origin O and hence O is the midpoint of every chord of the ellipse through O. Therefore the origin is the centre of the ellipse.
• Put y = 0 in the equation of the ellipse ⇒ x = a ⇒ x = ±a.
  Thus the curve meets x-axis (Principal axis) at two points A(a, 0), A'(-a, 0). Hence the ellipse has two vertices. The axis AA’ is called major axis. The length of the major axis is AA’ = 2a
• Put x = 0 ⇒ y² = b² ⇒ y = ± b. Thus, the curve meets y-axis (another axis) at two points B(0, b), B'(0, -b). The axis BB’ is called minor axis and the length of the minor axis is BB’ = 2b.
• The focus of the ellipse is S(ae, 0). The image of S with respect to the minor axis is S'(-ae,0). The point S’ is called second focus of the ellipse.
• The directrix of the ellipse is x = a/e. The image of x = a/e with respect to the minor axis is x = -a/e. The line x = -a/e is called second directrix of the ellipse.
• \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
  y² = b²(1 – \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\)) ⇒ y = \(\frac{b}{a} \sqrt{a^{2}-x^{2}}\)

Thus y has real values only when -a ≤ x ≤ a. Similarly x has real values only when -b ≤ y ≤ b. Thus the curve lies completely with in the rectangle x = ±a, y = ±b. Therefore the ellipse is a closed curve.

Theorem:
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (a > b > 0) is \(\frac{2 b^{2}}{a}\)
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (0 < a < b) is \(\frac{2 b^{2}}{a}\)

Let LL’ be the length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Focus S = (ae, 0)
If SL = 1, then L = (ae, 1)
L is lies on the ellipse ⇒ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
⇒ e + \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 ⇒ \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 – e = \(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\) ⇒ 1 = \(\frac{b^{4}}{a^{2}}\)
⇒ 1 = \(\frac{b^{2}}{a}\) ⇒ SL = \(\frac{b^{2}}{a}\)
LL’ = 2SL = \(\frac{2 b^{2}}{a}\)

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (a < b < 0) are L = (ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L’ = (ae, –\(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L₁ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)); L₁‘ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\))

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (0 < a < b) are L = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L’ = (-\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L₁ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be), L₁‘ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be)

Theorem:
If P is a point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 with foci S and S then PS + PS’ = 2a.
Proof:
Let e be the eccentricity and L = 0, L’ = 0 be the directrices of the ellipse.
Let C be the centre and A, A’ be the vertices of the ellipse.
∴ AA’ = 2a
Foci of the ellipse are S(ae, 0), S'(-ae, 0)
Let P(x₁, y₁) be a point on the ellipse

Let M, M’ be the projections of P on the directrices L = 0, L’ = 0 respectively.
∴ \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, \(\frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}\) = e

Let Z, Z’ be the points of intersection of major axis with directrices.
∴ MM’ = ZZ’ = CZ + CZ’ = 2a/e.
PS + PS’ = ePM + ePM’
= e(PM + PM’) = e(MM’) = e(2a/e) = 2a.

Theorem:
Let P(x₁, y₁) be a point and S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0 be an ellipse. Then
(i) P lies on the ellipse ⇔ S₁₁ = 0,
(ii) P lies inside the ellipse ⇔ S₁₁ < 0,
(iii) P lies outside the ellipse ⇔ S₁₁ > 0

Theorem:
The equation of the tangent to the ellipse S = 0 at F(x₁, y₁) is S₁ = 0.

Theorem:
The equation of the normal to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 F(x₁, y₁) is \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b².
Proof:
The equation of the tangent to S = 0 at F is S₁ = 0
⇒ \(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 =
The equation of the normal to S = 0 at F is
\(\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}}\)(x – x₁) – \(\frac{\mathrm{x}_{1}}{\mathrm{~a}^{2}}\)(y – y₁) = 0
⇒ \(\frac{x_{1}}{b^{2}}-\frac{y_{1}}{a^{2}}=\frac{x_{1} y_{1}}{b^{2}}-\frac{x_{1} y_{1}}{a^{2}}\)
⇒ \(\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{xy}}{\mathrm{b}^{2}}-\frac{\mathrm{y} \mathrm{x}_{1}}{\mathrm{a}^{2}}\right)=\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{~b}^{2}}-\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{a}^{2}}\right)\)
⇒ \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b²

Theorem:
The condition that the line y = mx + c may be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 is c² = a²m² + b².
Proof:
Suppose y = mx + c … (1)is a tangent to the e11ipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Let P(x1, y) be the point of contact.
The equation of the tangent at P is
\(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 = 0 … (2)

Now (1) and (2) represent the same line.
\(\frac{\mathrm{x}_{1}}{\mathrm{a}^{2} \mathrm{~m}}=\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}(-1)}=\frac{-1}{\mathrm{c}}\) ⇒ x₁ = \(\frac{-a^{2} m}{c}\), y₁ = \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\) = m\(\left(\frac{-a^{2} m}{c}\right)\) + c ⇒ b² = -a²m² + c²
⇒ c² = a²m² + b²

Note:
The equation of a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 may be taken as y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\). The point of contact is \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\) where c² = a²m² + b²

Theorem:
Two tangents can be drawn to an ellipse from an external point.

Director Circle:
The points of intersection of perpendicular tangents to an ellipse S = 0 lies on a circle, concentric with the ellipse.
Proof:
Equation of the ellipse
S ≡\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 0
Let P(x₁, y₁) be the point of intersection of perpendicular tangents drawn to ellipse.

Let y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) be a tangent to the ellipse S = 0 passing through P.
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ y₁ – mx₁ = ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ (y₁ – mx₁)² = a²m² + b²
⇒ y₁² + m²x₁² – 2x₁y₁m = a²m² + b²
⇒ (x₁² – a²)m² -2x₁y₁m + (y₁² -b²) = 0 … (1)

If m₁, m₂ are the slopes of the tangents through P then m1, m2 are the roots of (1).
The tangents through P are perpendicular.
⇒ m₁m₂ = -1 ⇒ \(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\) = -1
⇒ y₁² – b² = -x² + a² ⇒ x²₁ + y²₁ = a² + b²
∴ P lies on x² + y² = a² + b² which is a circle with centre as origin, the centre of the ellipse.

Auxiliary Circle:
Theorem: The feet of the perpendiculars drawn from either of the foci to any tangent to the ellipse S = 0 lies on a circle, concentric with the ellipse.( called auxiliary circle)
Proof:
Equation of the ellipse S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0
Let P(x₁, y₁) be the foot of the perpendicular drawn from either of the foci to a tangent.
The equation of the tangent to the ellipse S = 0 is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) … (1)
The equation to the perpendicular from either foci (±ae, 0) on this tangent is
y = –\(\frac{1}{m}\)(x ± ae)

Now P is the point of intersection of (1) and (2)

∴ y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\), y₁ = –\(\frac{1}{m}\)(x₁ ± ae
⇒ y₁ – mx₁ = ±V a2m2 + b2, my₁ + x₁ = ±ae
⇒ (y₁ – mx₁)² + (my₁ + x₁)² = a²m² + b² + a²e²
⇒ y₁² + m²x² – 2x₁y₁m + m²y₁² + x₁² + 2x₁y₁m = a²m² + a²(1 – e²) + a²e²
⇒ x₁²(m² + 1) + y₁²(1 + m²) = a²m² + a²
⇒ (x₁² + y₁²)(m² + 1) = a² (m² + 1)
⇒ x₁² + y₁² = a²
P lies on x² + y² = a² which is a circle with centre as origin, the centre of the ellipse.

Theorem:
The equation to the chord of contact of P(x₁, y₁) with respect to the ellipse S = 0 is S₁ = 0.

Eccentric Angle Definition:
Let P(x, y) be a point on the ellipse with centre C. Let N be the foot of the perpendicular of P on the major axis. Let NP meets the auxiliary circle at P’. Then ∠NCP’ is called eccentric angle of P. The point P’ is called the corresponding point of P.

Parametric Equations: If P(x, y) is a point on the ellipse then x = a cos θ, y = b sin θ where θ is the eccentric angle of P. These equations x = a cos θ, y = b sin θ are called parametric equations of the ellipse. The point P(a cos θ, b sin θ) is simply denoted by θ.

Theorem: The equation of the chord joining the points with eccentric angles α and β on the ellipse S = 0 is \(\frac{x}{a} \cos \frac{\alpha+\beta}{2}+\frac{y}{b} \sin \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2}\)
Proof:
Given points on the ellipse are P(a cos α, b sin α), Q(a cos β, b sin β).
Slope of \(\overline{\mathrm{PQ}}\) is \(\frac{\mathrm{b} \sin \alpha-\mathrm{b} \sin \beta}{\mathrm{a} \cos \alpha-\mathrm{a} \cos \beta}=\frac{\mathrm{b}(\sin \alpha-\sin \beta)}{\mathrm{a}(\cos \alpha-\cos \beta)}\)
Equation of \(\overline{\mathrm{PQ}}\) is:

Theorem: The equation of the tangent at P(θ) on the ellipse
S = 0 is \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1.

Theorem: The Equation of The Normal At P(θ) On The Ellipse
S = 0 Is \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}\) = a² – b².

Theorem: Four normals can be drawn from any point to the ellipse and the sum of the eccentric angles of their feet is an odd multiple of π.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 3 Parabola to solve questions creatively.

Intermediate 2nd Year Maths 2B Parabola Formulas

Definition:
→ Conic section: If right circular solid cone is cut by a plane the section of it is called a Conic section.

→ Conic: The locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ is called a conic.

→ Parabola : A conic with eccentricity 7 is called a parabola i.e., the locus of a point, whose distance from fixed point (focus) is equal to the distance from a fixed straight line (directrix) is called a parabola.

→ Axis of Parabola: The line passing through the vertex and the focus and perpendicular to directrix of the parabola is called the axis of the Parabola.

→ Equation of Parabola : General form – Let S(α, β) be the focus and ax + by + c = 0 be directrix then by definition the equation of parabola be,
\(\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)

Various forms of parabola:
→ y² = 4ax
Axis : X – axis
Focus : (a, 0)
Vertex : (0, 0)
Equation of directrix : x + a = 0

→ y²= – 4ax
Axis: x – axis
Focus: (- a, 0)
Vertex : (0, 0)
Equation of directrix : x – a = 0

→ x² = 4ay (a > 0)
Axis: Y-axis Focus : (0, a)
Vertex : (0, 0)
Equation of directrix: y + a = 0

→ x² = -4ay(a>0)
Axis : Y axis
Focus : (0, – a)
Vertex: (0, 0)
Equation of directrix : y – a = 0

→ (y – k)² = 4a (x – h) a > 0
Axis : y = k’ a line parallel to X- axis
Focus : (a + h, k)
Vertex: (h, k)
Equation of directrix: x + a = h

→ (x – h)² = 4a (y – k) a > 0
Axis: x = h a line parallel to Y-axis
Focus: (h, a + k)
Vertex: (h, k)
Equation of directrix: y + a = k

→ \(\sqrt{(x-h)^{2}+(y-k)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)
Axis: b(x – h) – a(y – k) = 0
Focus: (h, k)
Vertex: Some point A(fig)
Equation of directrix : ax + by + c = 0

Chord: Line segment joining two points of a parabola is called a, chord of the parabola. If this chord passes through focus is called focal chord.
Chord passing through focus and ⊥ to axis is called latus rectum.
Length of latus rectum = 4a

Parametric form of parabola:
For y² = 4ax
Parametric form will be x = at², y = 2at

Definition:
→ The locus of a point which moves in a plane so that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line is called a conic section or conic. The fixed point is called focus, the fixed straight line is called directrix and the constant ratio ‘e’ is called eccentricity of the conic.

If e = 1, then the conic is called a Parabola.
If e < 1, then the conic is called an Ellipse. If e > 1, then the conic is called a hyperbola.
Note: The equation of a conic is of the form ax² + 2hxy + by² + 2gx + 2fy + c = 0.

→ Directrix of the Conic: A line L = 0 passing through the focus of a conic is said to be the principal axis of the conic if it is perpendicular to the directrix of the conic.

→ Vertices: The points of intersection of a conic and its principal axis are called vertices of the conic.

→ Centre: The midpoint o the line segment joining the vertices of a conic is called centre of the conic.

→ Note 1: If a conic has only one vertex then its centre coincides with the vertex.

→ Note 2: If a conic has two vertices then its centre does not coincide either of the vertices. In this case the conic is called a central conic.

→ Standard Form: A conic is said to be in the standard form if the principal axis of the conic is x-axis and the centre of the conic is the origin.

Equation of a Parabola in Standard Form:
The equation of a parabola in the standard form is y² = 4ax.
Proof :
Let S be the focus and L = 0 be the directrix of the parabola.
Let P be a point on the parabola.
Let M, Z be the projections of P, S on the directrix L = 0 respectively.
Let N be the projection of P on SZ.
Let A be the midpoint of SZ.
Therefore, SA = AZ, ⇒ A lies on the parabola. Let AS = a.
Let AS, the principal axis of the parabola as x-axis and Ay perpendicular to SZ as y-axis.
Then S = (a, 0) and the parabola is in the standard form.
Let P = (x₁, y₁).

Now PM = NZ = NA + AZ = x₁ + a
P lies on the parabola ⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = 1 ⇒ PS = PM
⇒ \(\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-0\right)^{2}}\) = x₁ + a
⇒ (x₁ – a)² + y₁² = (x₁ + a)²
⇒ y₁² = (x₁ + a)² – (x₁ – a)²
⇒ y₁² = 4ax₁
The locus of P is y² = 4ax.
∴ The equation to the parabola is y² = 4ax.

Nature of the Curve y² = 4ax.

(i) The curve is symmetric with respect to the x-axis.
∴ The principal axis (x-axis) is an axis of the parabola.

(ii) y = 0 ⇒ x = 0. Thus the curve meets x-axis at only one point (0, 0).
Hence the parabola has only one vertex.

(iii) If x < 0 then there exists no y ∈ R. Thus the parabola does not lie in the second and third quadrants. (iv) If x > 0 then y² > 0 and hence y has two real values (positive and negative). Thus the parabola lies in the first and fourth quadrants.

(v) x = 0 ⇒ y² = 0 ⇒ y = 0, 0. Thus y-axis meets the parabola in two coincident points and hence y-axis touches the parabola at (0, 0).

(vi) As x → ∞ ⇒ y² → ∞ ⇒ y → ±∞
Thus the curve is not bounded (closed) on the right side of the y-axis.

→ Double Ordinate: A chord passing through a point P on the parabola and perpendicular to the principal axis of the parabola is called the double ordinate of the point P.

→ Focal Chord: A chord of the parabola passing through the focus is called a focal chord.

→ Latus Rectum: A focal chord of a parabola perpendicular to the principal axis of the parabola is called latus rectum. If the latus rectum meets the parabola in L and L’, then LL’ is called length of the latus rectum.

Theorem: The length of the latus rectum of the parabola y² = 4ax is 4a.
Proof:
Let LL’ be the length of the latus rectum of the parabola y² = 4ax.

Let SL = 1, then L = (a, 1)
Since L is a point on the parabola y² = 4ax, therefore 1² = 4a(a)
⇒ 1² = 4a² ⇒ 1 = 2a ⇒ SL = 2a
∴ LL’ = 2SL = 4a.

→ Focal Distance: If P is a point on the parabola with focus S, then SP is called focal distance of P.

Theorem: The focal distance of P(x₁, y₁) on the parabola y² = 4ax is x₁ + a.
Notation: We use the following notation in this chapter
S ≡ y² – 4ax
S₁ ≡ YY₁ – 2a(x + x₁)
S₁₁ = S(x₁, y₁) ≡ y₁² – 4ax₁
S₁₂ ≡ y₁y₂ – 2a(x₁ + x₂)

Note:
Let P(x₁, y₁) be a point and S ≡ y² – 4ax = 0 be a parabola. Then

• P lies on the parabola ⇔ S₁₁ = 0
• P lies inside the parabola ⇔ S₁₁ = 0
• P lies outside the parabola ⇔ S₁₁ = 0

Theorem: The equation of the chord joining the two points A(x₁, y₁), B(x₂, y₂) on the parabola S = 0 is S₁ + S₂ = S₁₂.
Theorem: The equation of the tangent to the parabola S = 0 at P(x₁, y₁) is S₁ = 0.

Normal:
Let S = 0 be a parabola and P be a point on the parabola S = 0. The line passing through P and perpendicular to the tangent of S = 0 at P is called the normal to the parabola S = 0 at P.

Theorem: The equation of the normal to the parabola y² = 4ax at P(x₁, y₁) is y₁(x – x₁) + 2a(y – y₁) = 0.
Proof:
The equation of the tangent to S = 0 at P is S₁ = 0

⇒ yy₁ – 2a(x + x₁) = 0.
⇒ yy₁ – 2ax – 2ax₁ = 0
The equation of the normal to S = 0 at P is:
y₁(x – x₁) + 2a(y – y₁) = 0

Theorem: The condition that the line y = mx + c may be a tangent to the parabola y² = 4ax is c = a/m.
Proof:
Equation of the parabola is y² = 4ax ……….. (1)
Equation of the line is y = mx + c ………. (2)
Solving (1) and (2),
(mx + c)² = 4ax ⇒ m²x² + c² + 2mcx = 4ax
⇒ m²x² + 2(mc – 2a)x + c² = 0
Which is a quadratic equation in x. Therefore it has two roots.
If (2) is a tangent to the parabola, then the roots of the above equation are equal.
⇒ its disc eminent is zero
⇒ 4(mc – 2a)² – 4m²c² = 0
⇒ m²c² + 4a² – 4amc – m²c² = 0
⇒ a² – amc = 0
⇒ a = mc
⇒ C = \(\frac{a}{m}\)

II Method:
Given parabola is y² = 4ax.
Equation of the tangent is y = mx + c ———— (1)
Let P(x₁, y₁) be the point of contact.
The equation of the tangent at P is
yy₁ – 2a(x + x₁) = 0 ⇒ yy₁ = 2ax + 2ax₁ ……. (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{y}_{1}}{1}=\frac{2 \mathrm{a}}{\mathrm{m}}=\frac{2 \mathrm{ax}_{1}}{\mathrm{c}}\) ⇒ x₁ = \(\frac{\mathrm{c}}{\mathrm{m}}\), y₁ = \(\frac{\mathrm{2a}}{\mathrm{m}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{2 \mathrm{a}}{\mathrm{m}}=\mathrm{m}\left(\frac{\mathrm{c}}{\mathrm{m}}\right)+\mathrm{c} \Rightarrow \frac{2 \mathrm{a}}{\mathrm{m}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)

Note: The equation of a tangent to the parabola y² = 4ax can be taken as y = mx + a/m. And the point of contact is (a/m², 2a/m).

Corollary: The condition that the line lx + my + n = 0 to touch the parabola y² = 4ax is am² = ln.
Proof:
Equation of the parabola is y² = 4ax …………. (1)
Equation of the line is lx + my + n = 0
⇒ y = – \(\frac{1}{\mathrm{~m}}\)x – \(\frac{\mathrm{n}}{\mathrm{m}}\)
But this line is a tangent to the parabola, therefore
C = a/m ⇒ \(-\frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{a}}{-1 / \mathrm{m}} \Rightarrow \frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{am}}{1}\) ⇒ am² = ln
Hence the condition that the line lx + my + n = 0 to touché the parabola y² = 4ax is am² = ln.

Note: The point of contact of lx + my + n = 0 with y² = 4ax is (n/l, – 2am/l).

Theorem: The condition that the line lx + my + n = 0 to touch the parabola x² = 4ax is al² = mn.
Proof:
Given line is lx + my + n = 0 …… (1)
Let P(x₁, y₁) be the point of contact of (1) with the parabola x² = 4ay.
The equation of the tangent at P to the parabola is xx₁ = 2a(y + y₁)
⇒ x₁x – 2ay – 2ay₁ = 0 …… (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{x}_{1}}{1}=-\frac{2 \mathrm{a}}{\mathrm{m}}=-\frac{2 \mathrm{ay}_{1}}{\mathrm{n}}\) ⇒ x₁ = – \(-\frac{2 \mathrm{al}}{\mathrm{m}}\), y₁ = \(\frac{n}{m}\)
P lies on the line lx + my + n = 0
⇒ lx₁ + my₁ + n = 0 ⇒ l\(\left(\frac{-2 a l}{m}\right)\) + m\(\left(\frac{\mathrm{n}}{\mathrm{m}}\right)\) + n = 0
⇒ – 2al² + mn + mn = 0 ⇒ al² = mn

Theorem: Two tangents can be drawn to a parabola from an external point.

Note:
1. If m₁, m₂ are the slopes of the tangents through P, then m₁, m₂ become the roots of
equation (1). Hence m₁ + m₂ = y₁/x₁, m₁m₂ = a/x₁.

2. If P is a point on the parabola S =0 then the roots of equation (1) coincide and hence only one tangent can be drawn to the parabola through P.

3. If P is an internal point to the parabola S = 0 then the roots of (1) are imaginary and hence no tangent can be drawn to the parabola through P.

Theorem: The equation in the chord of contact of P(x₁, y₁) with respect to the parabola S = 0 is S₁ = 0.

Theorem: The equation of the chord of the parabola S = 0 having P(x₁, y₁) as its midpoint is S₁ = S₁₁.

Pair of Tangents:
Theorem: The equation to the pair of tangents to the parabola S = 0 from P(x₁, y₁) is S₁² = s₁₁s.

Parametric Equations of the Parabola:
A point (x, y) on the parabola y² = 4ax can be represented as x = at², y = 2at in a single parameter t. Theses equations are called parametric equations of the parabola y² = 4ax. The point (at², 2at) is simply denoted by t.

Theorem: The equation of the tangent at (at², 2at) to the parabola is y² = 4ax is yt = x + at².
Proof:
Equation of the parabola is y² = 4ax.
Equation of the tangent at (at², 2at) is S₁ = 0.
⇒ (2at)y – 2a(x + at²) = 0
⇒ 2aty = 2a(x + at²) ⇒ yt = x + at².

Theorem: The equation of the normal to the parabola y² = 4ax at the point t is y + xt = 2at + at³.
Proof:
Equation of the parabola is y² = 4ax.
The equation of the tangent at t is:
yt = x + at² = x – yt + at² = 0
The equation of the normal at (at², 2at) is
t(x – at²) + l(y – 2at) = 0
⇒ xt – at³ + y – 2at = 0 ⇒ y + xt = 2at + at³

Theorem: The equation of the chord joining the points t₁ and t₂ on the parabola y² = 4ax is y(t₁ + t₂) = 2x + 2at₁t₂.
Proof:
Equation of the parabola is y² = 4ax.
Given points on the parabola are
P(at₁², 2at₁), Q(at₂², 2at₂) .
Slope of \(\overline{\mathrm{PQ}}\) is
\(\frac{2 \mathrm{at}_{2}-2 \mathrm{at}_{1}}{\mathrm{at}_{2}^{2}-\mathrm{at}_{1}^{2}}=\frac{2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{a}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right)}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}\)
The equation of \(\overline{\mathrm{PQ}}\) is y – 2at₁ = \(\frac{2}{t_{1}+t_{2}}\) (x – at₁²).
⇒ (y – 2at₁) (t₁ + t₂) = 2(x – at₁²)
⇒ y(t₁ + t₂) – 2at₁² – 2at₁t₂ = 2x – 2at₁²
⇒ y(t₁ + t₂) = 2x + 2at₁t₂.

Note: If the chord joining the points t₁ and t₂ on the parabola y² = 4ax is a focal chord then t₁t₂ = – 1.
Proof:
Equation of the parabola is y² = 4ax
Focus S = (a, o)
The equation of the chord is y(t₁ + t₂) = 2x + 2at₁t₂
If this is a focal chord then it passes through the focus (a, 0).
∴ 0 = 2a + 2at₁t₂ ⇒ t₁t₂ = – 1.

Theorem: The point of intersection of the tangents to the parabola y² = 4ax at the
points t₁ and t₂ is (at₁t₂, a[t₂ + t₂]).
Proof:
Equation of the parabola is y² = 4ax
The equation of the tangent at t₁ is yt₁ = x + at₁² ……. (1)
The equation of the tangent at t₂ is
yt₂ = x + at₂² ……….. (2)
(1) – (2) ⇒ y(t₁ – t₂) = a(t₁² – t₂²) ⇒ y = a(t₁ + t₂)
(1) ⇒ a(t₁ + t₂)t₁ = x + at₁²
= at₁² + at₁t₁ = x + at₁² ⇒ x = at₁t₂,
∴ Point of intersection = (at₁t₂, a[t₁ + t₂]).

Theorem: Three normals can be drawn form a point (x₁, y₁) to the parabola y² = 4ax.
Corollary: If the normal at t₁ and t₂, to the parabola y² = 4ax meet on the parabola, then t₁t₂ = 2.
Proof:
Let the normals at t₁ and t₂ meet at t₃ on the parabola.
The equation of the normal at t1 is:
y + xt₁ = 2at₁ + at₁³ ………… (1)
Equation of the chord joining t1 and t3 is:
y(t₁ + t₃) = 2x + 2at₁t₃ ……… (2)

(1) and (2) represent the same line
∴ \(\frac{t_{1}+t_{3}}{1}=\frac{-2}{t_{1}} \Rightarrow t_{3}=-t_{1}-\frac{2}{t_{1}}\)
Similarly t₃ = – t₁ – \(\frac{2}{\mathrm{t}_{2}}\)
∴ \(-\mathrm{t}_{1}-\frac{2}{\mathrm{t}_{1}}=-\mathrm{t}_{2}-\frac{2}{\mathrm{t}_{2}}\) ⇒ t₁ – t₂ \(\frac{2}{\mathrm{t}_{2}}-\frac{2}{\mathrm{t}_{1}}\)
⇒ t₁ – t₂ = \(\frac{2\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}{\mathrm{t}_{1} \mathrm{t}_{2}}\) ⇒ t₁t₂ = 2

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 2 System of Circles to solve questions creatively.

Intermediate 2nd Year Maths 2B System of Circles Formulas

Definition:
→ The angle between two intersecting circles is defined as the angle between the tangents at the point of intersection of the two circle’s. If θ is the angle between the circles, then
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
Here d = distance between the centres, r₁, r₂ be their radii.

→ If θ is angle between the two circles.
x² + y² + 2gx + 2fy + c = 0 and x² + y² + 2g’x + 2f’y + c’ = 0, then
cos θ = \(\frac{-2 g g^{\prime}-2 f f^{\prime}+c+c^{\prime}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

→ Circles cut orthogonally if
2g’g + 2f’f = c + c’ [∵ cos 90° = 0]
(or) d² = r²₁ + r²₂ then also two circles cut orthogonally.

Theorem:
If d is the distance between the centers of two intersecting circles with radii r₁, r₂ and θ is the angle between the circles then cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\).

Proof:
Let C₁, C₂ be the centre s of the two circles S = 0, S’ = 0 with radii r₁, r₂ respectively. Thus C₁C₂ = d. Let P be a point of intersection of the two circles. Let PB, PA be the tangents of the circles S = 0, S’ = 0 respectively at P.

Now PC₁ = r₁, PC₂ = r₂, ∠APB = θ
Since PB is a tangent to the circle S = 0, ∠C₁PB = π/2
Since PA is a tangent to the circle S’ = 0, ∠C₂PA = π/2
Now ∠C₁PC₂ = ∠C₁PB + ∠C₂PA – ∠APB = π/2 + π/2 – θ = π – θ
From ∆C₁PC₂, by cosine rule,
C₁²C₂² = PC₁² + PC₂² – 2PC₁ . PC₂ cos ∠C₁PC₂ ⇒ d² = r₁² + r₂² – 2r₁r₂ cos(π – θ) ⇒ d² = r₁² + r₂² + 2r₁r₂ cos θ
⇒ 2r₁r₂ cos θ = d² – r₁² – r₂² ⇒ cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Corollary:
If θ is the angle between the circles x² + y² + 2gx + 2fy + c = 0, x² + y² + 2g’x + 2f’y + c’= 0 then cos θ = \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
proof:
Let C₁, C₂ be the centre s and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively and C₁C₂ = d.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’),
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c}\)
Now cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\) = \(\frac{\left(g-g^{\prime}\right)^{2}+\left(f-f^{\prime}\right)^{2}-\left(g^{2}+f^{2}-c\right)-\left(g^{\prime 2}+f^{\prime 2}-c^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
= \(\frac{\mathrm{g}^{2}+\mathrm{g}^{\prime 2}-2 \mathrm{gg}^{\prime}+\mathrm{f}^{2}+\mathrm{f}^{\prime 2}-2 \mathrm{ff} \mathrm{f}^{\prime}-\mathrm{g}^{2}-\mathrm{f}^{2}+\mathrm{c}-\mathrm{g}^{2}-\mathrm{f}^{\prime 2}+\mathrm{c}^{\prime}}{2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}} \sqrt{\mathrm{g}^{2}+\mathrm{f}^{\prime 2}-\mathrm{c}^{\prime}}}\)
= \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

Note: Let d be the distance between the centers of two intersecting circles with radii r₁, r₂. The two circles cut orthogonally if d² = r₁² + r₂².
Note: The condition that the two circles

S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’ = 0 may cut each other orthogonally is 2gg’ + 2ff’ = c + c’.
Proof: Let C₁, C₂ be the centers and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’)
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c^{\prime}}\)
Let P be point of intersection of the circles.
The two circles cut orthogonally at P
⇔ ∠C₁PC₂ = 90°.
⇒ C₁C₂² = C₁P² + C₂P² ⇔ (g – g’)² + (f – f’)² = r₁² + r₂²;
⇔ g² + g’² – 2gg’ + f² + f’² – 2ff’ = g² + f² – c + g’² + f’² + c’
⇔ – (2gg’ + 2ff’) = – (c + c’) ⇒ 2gg’+ 2ff’ = c + c’

Note:

• The equation of the common chord of the intersecting circles s = 0 and s¹ = 0 is s – s¹ = 0.
• The equation of the common tangent of the touching circles s = 0 and s¹ = 0 is s – s¹ = 0
• If the circle s = 0 and the line L = 0 are intersecting then the equation of the circle passing through the points of intersection of s = 0 and L = 0 is S + λL = 0.
• The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is s + λS’ = 0.

Theorem: The equation of the radical axis of the circles S = 0, S’ = 0 is S – S’ = 0.

Theorem: The radical axis of two circles is perpendicular to their line of centers.
Proof:
Let S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’= 0 be the given circles.

The equation of the radical axis is S – S’ = 0
⇒ 2(g – g’)x + 2(f – f’)y + (c – c’) = 0
⇒ a₁x + b₁y + c₁ = 0 where
a₁ = 2(g – g’), b₁ = 2(f – f’), c₁ = e – e’
The centers of the circles are (- g, – f), (- g’, – f’)
The equation to the line of centers is:
(x + g) (f – f’) = (y + f) (g – g’)
⇒ (f – f’)x – (g – g’)y – gf’ + fg’= 0
⇒ a₂x + b₂y + c₂ = 0 where
a₂ = f – f’, b₂ = – (g – g’), c₂ = fg’ – gf’
Now a₁a₂ + b₁b₂ = 2(g – g’) (f – f’) – 2(f – f’) (g – g’) = 0.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 1 Circle to solve questions creatively.

Intermediate 2nd Year Maths 2B Circle Formulas

→ The locus of a point in a plane such that its distance from a fixed point in the plane is always the same is called a circle.

→ The equation of a circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r²

→ The equation of a circle in standard form is x² + y² = r².

→ The equation of a circle in general form is x² + y² + 2gx + 2fy + c = 0 and its centre is (-g, -f), radius is \(\sqrt{g^{2}+f^{2}-c}\).

→ The intercept made by x² + y² + 2gx + 2fy + c = 0

• on X-axis is 2\(\sqrt{g^{2}-c}\) if g² > c.
• on Y-axis is 2\(\sqrt{f^{2}-c}\) if f² > c.

→ If the extremities of a diameter of a circle are (x₁, y₁) and (x₂, y₂) then its equation is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

→ The equation of a circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\) = (x² + y²) + \(\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|\) x + \(\left|\begin{array}{lll}
x_{1} & C_{1} & 1 \\
x_{2} & C_{2} & 1 \\
x_{3} & C_{3} & 1
\end{array}\right|\) y + \(\left|\begin{array}{lll}
x_{1} & y_{1} & C_{1} \\
x_{2} & y_{2} & C_{2} \\
x_{3} & y_{3} & C_{3}
\end{array}\right|\) = 0.
where c_i = – (x_i² + y_i²)

→ The centre of the circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left[\frac{\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}, \frac{\left|\begin{array}{lll}
x_{1} & c_{1} & 1 \\
x_{2} & c_{2} & 1 \\
x_{3} & c_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}\right]\)

→ The parametric equations of a circle with centre (h, k) and radius (r ≥ 0) are given by
x = h + r cos θ
y = k + r sin θ 0 ≤ 6 < 2π.

→ A point P(x₁, y₁) is an interior point or on the circumference or an exterior point of a circles S = 0 ⇔ S₁₁ \(\frac{<}{>}\) 0.

→ The power of P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

→ A point P(x₁, y₁) is an interior point or on the circumference or exterior point of the circle S = 0 ⇔ the power of P with respect to S = 0 is negative, zero and positive.

→ If a straight line through a point P(x₁, y₁) meets the circle S = 0 at A and B then the power of P is equal to PA. PB.

→ The length of the tangent from P(x₁, y₁) to S = 0 is \(\sqrt{S_{11}}\).

→ The straight line l = 0 intersects, touches or does not meet the circles = 0 according as l < r, l = r or l > r where l is the perpendicular distance from the centre of the circle to the line l = 0 and r is the radius.

→ For every real value of m the straight line y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x² + y² = r².

→ If r is the radius of the circle S = x² + y² + 2gx + 2fy + c = 0 then for every real value of m the straight line y + f = m(n + g) ±r + m2 will be a tangent to the circle.

→ If P(x₁, y₁) and Q(x₂, y₂) are two points on the circle S = 0 then the secant’s \((\stackrel{\leftrightarrow}{P Q})\) equation is S₁ + S₂ = S₁₂

→ The equation of tangent at (x₁, y₁) of the circle S = 0 is S₁ = 0.

→ If θ₁, θ₂ are two points on S = x² + y² + 2gx + 2fy + c = 0 then the equation of the chord joining the points θ₁, θ₂ is
(x + g) cos \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) + (y + f) sin \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) = r cos \(\left(\frac{\theta_{1}-\theta_{2}}{2}\right)\)

→ The equation of the tangent at θ of the circle S = 0 is (x + g) cos θ + (y + f) sin θ = r.

→ The equation of normal at (x₁, y₁) of the circle
S = 0 is (x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0.

→ The chord of contact of P(x₁ y₁) (exterior point) with respect to S = 0 is S₁ = 0.

→ The equation of the polar of a point P(x₁, y₁) with respect to S = 0 is S₁ = 0.

P(x1, y1)	Tangent at P	Chord of contact at p	Polar of P
(i) Interior of the circle	Does not exist	Does not exist (not defined)	S1 = 0 (P is different from the centre of the circle)
(ii) On the circle	S1 = 0	S1 = 0	S1 = 0
(iii) Exterior of the circle	Does not exist	S1 = 0	S1 = 0

→ The pole of lx + my + n = 0 with respect to S = 0 is
\(\left(-g+\frac{l r^{2}}{l g+m f-n},-f+\frac{m r^{2}}{l g+m f-n}\right)\)

→ Where r is the radius of the circle. The polar of P(x₁, y₁) with respect to S = 0 passes through Q(x₂, y₂) ⇔ the polar of Q with respect to S – 0 passes through P.

→ The points (x₁, y₁) and (x₂, y₂) are conjugate points with respect to S = 0 if S₁₂ = 0

→ Two lines l₁x = m₁y + n₁ = 0, l₂x + m₂y + n₂ = 0 are conjugate with respect to x² + y² = a² ⇔ (l₁l₂ + m₁m₂) = n₁n₂

→ Two points P, Q are said to be inverse points with respect to S = 0 if CP. CQ = r² where C is the centre and r is the radius of the circle S = 0.

→ If (x₁, y₁) is the mid-point of a chord of the circle S = 0 then its chord equation is S₁ = S₁₁.

→ The pair of common tangents to the circles S = 0, S’ = 0 touching at a point on the lines segment \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) (C₁, C₂ are centres of the circles) is called transverse pair of common tangents.

→ The pair of common tangents to the circles S = 0, S’ = 0 intersecting at a point not in \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) is called as direct pair of common tangents.

→ The point of intersection of transvese (direct) common tangents is called internal (external) Centre of similitude.

Situation	No of common tangents
1. \( \overline{C_{1} C_{2}} \) > r1 + r2	4
2. r1 + r2 = \( \overline{C_{1} C_{2}} \)	3
3. |r1 – r2| < \( \overline{C_{1} C_{2}} \) < r1 + r2	2
4. C1C2 = |r1 – r2|	1
5. C1C2 < |r1 – r2|	0

→ The combined equation of the pair of tangents drawn from an external point P(x₁, y₁) to the circle S = 0 is SS₁₁ = S²₁.

Equation of a Circle:
The equation of the circle with centre C (h, k) and radius r is (x – h)² + (y – k)² = r².
Proof:
Let P(x₁, y₁) be a point on the circle.
P lies in the circle ⇔ PC = r ⇔ \(\sqrt{\left(\mathrm{x}_{1}-\mathrm{h}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{k}\right)^{2}}\) = r
⇔ (x₁ – h)² + (y₁ – k)² = r².

The locus of P is (x – h)² + (y – k)² = r².
∴ The equation of the circle is (x – h)² + (y – k)² = r².

Note: The equation of a circle with centre origin and radius r is (x – 0)² + (y – 0)² = r²
i.e., x² + y² = r² which is the standard equation of the circle.

Note: On expanding equation (1), the equation of a circle is of the form x² + y² + 2gx + 2fy + c = 0.

Theorem: If g² + f² – c ≥ 0, then the equation x² + y² + 2gx + 2fy + c = 0 represents a circle with centre (- g, – f) and radius \(\sqrt{g^{2}+f^{2}-c}\).
Note: If ax² + ay² + 2gx + 2fy + c = 0 represents a circle, then its centre = \(\left(-\frac{g}{a},-\frac{f}{a}\right)\) and its radius \(\frac{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{ac}}}{|\mathrm{a}|}\).

Theorem: The equation of a circle having the line segment joining A(x₁, y₁) and B(x₂, y₂) as diameter is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0.

Let P(x, y) be any point on the circle. Given points A(x₁, y₁) and B(x₂, y₂).
Now ∠APB = \(\frac{\pi}{2}\). (Angle in a semi circle.)
Slope of AP. Slope of BP = – 1
⇒ \(\frac{y-y_{1}}{x-x_{1}} \frac{y-y_{2}}{x-x_{2}}\) = – 1
⇒ (y – y₂) (y – y₁) = – (x – x₂) (x – x₁) = 0
⇒ (x – x₂) (x – x₁) + (y – y₂) (y – y₁) = 0

Definition: Two circles are said to be concentric if they have same center.

The equation of the circle concentric with the circle x² + y² + 2gx + 2fy + c = 0 is of the form x² + y² + 2gx + 2fy + k = 0.
The equation of the concentric circles differs by constant only.

Parametric Equations of A Circle:

Theorem: If P(x, y) is a point on the circle with centre C(α, β) and radius r, then x = α + r cosθ, y = β + r sin θ where 0 ≤ θ < 2π.

Note: The equations x = α + r cos θ, y = + r sin θ, 0 ≤ θ < 2π are called parametric equations of the circle with centre (α, β) and radius r.

Note: A point on the circle x² + y² = r² is taken in the form (r cos θ, r sin θ). The point (r cos θ, r sin θ) is simply denoted as point θ.

Theorem:
(1) If g² – c > 0 then the intercept made on the x axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{g^{2}-a c}\)
(2) If f² – c >0 then the intercept made on the y axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{f^{2}-b c}\)

Note: The condition for the x-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is g² = c.

Note: The condition of the y-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is f² = c.

Position of Point:
Let S = 0 be a circle and P(x₁, y₁) be a point I in the plane of the circle. Then

• P lies inside the circle S = 0 ⇔ S₁₁ < 0
• P lies in the circle S = 0 ⇔ S₁₁ = 0
• Plies outside the circle S = 0 ⇔ S₁₁ = 0

Power of a Point:
Let S = 0 be a circle with centre C and radius r. Let P be a point. Then CP² – r² is called power of P with respect to the circle S = 0.

Theorem: The power of a point P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

Theorem: The length of the tangent drawn from an external point P(x₁, y₁) to the circle s = 0 is \(\sqrt{\mathrm{S}_{11}}\).

Theorem: The equation of the tangent to the circle S = 0 at P(x₁, y₁) is S₁ = 0.

Theorem: The equation of the normal to the circle S = x² + y² + 2gx + 2fy + c = 0 at P(x₁, y₁) is
(y₁ + f) (x – x₁) – (x₁ + g) (y – y₁) = 0.

Corollary: The equation of the normal to the circle x² + y² = a² at P(x₁, y₁) is y₁x – x₁y = 0.

Theorem: The condition that the straight line lx + my + n = 0 may touch the circle x² + y² = a² is n² = a²(l² + m²) and the point of contact is \(\left(\frac{-a^{2} 1}{n}, \frac{-a^{2} m}{n}\right)\).
Proof:
The given line is lx + my + n = 0 …… (1)
The given circle is x² + y² = r² ……. (2)
Centre C = (0, 0), radius = r
Line (1) is a tangent to the circle (2)
⇔ The perpendicular distance from the centre C to the line (1) is equal to the radius r.
⇔ \(\left|\frac{0-n}{\sqrt{1^{2}+m^{2}}}\right|\) = r
⇔ (n)² = r² (l² + m²)

Let P(x₁, y₁) be the point of contact.
Equation of the tangent is S₁ = 0, ⇒ x₁x + y₁y – r² = 0. —- (3)
Equations (1) and (3) are representing the same line, therefore, \(\frac{x_{1}}{l}=\frac{y_{1}}{m}=\frac{-a^{2}}{n}\) ⇒ x₁ = \(\frac{-a^{2} l}{n}\), y₁ = \(\frac{-a^{2} m}{n}\)
Therefore, point of contact is \(\left(\frac{-a^{2} l}{n}, \frac{-a^{2} m}{n}\right)\)

Theorem: The condition for the straight line lx + my + n = 0 may be a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 is (g² + f² – c) (l² + m²) = (lg + mf – n)².
Proof:
The given line is lx + my + n = 0 …….. (1)
The given circle is x² + y² + 2gx + 2fy + c = 0 …….. (2)

Centre C = (- g, – f), radius r = \(\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)
Line (1) is a tangent to the circle (2)
⇔ The perpendicular distance from the centre C to the line (1) is equal to the radius r.
⇔ \(\left|\frac{-\lg -m f+c}{\sqrt{1^{2}+m^{2}}}\right|\) = \(\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\)
⇔ (lg + mf -n)² = (g² + f² – c) (l² + m²)

Corollary: The condition for the straight line y = mx + c to touch the circle
x² + y² = r² is c² = r²(1 + m²).
The given line is y = mx + c i.e., mx – y + c = 0 … (1)
The given circle is S = x² + y² = r²
Centre C = (0,0), radius = r.
If (1) is a tangent to the circle, then
Radius of the circle = perpendicular distance from centre of the circle to the line.

⇒ r = \(\frac{|c|}{\sqrt{m^{2}+1}}\) ⇒ r² = \(\frac{c^{2}}{m^{2}+1}\) ⇒ r² (m² + 1) = c²

Corollary: If the straight line y = mx + c touches the circle x² + y² = r², then their point of contact is \(\left(-\frac{r^{2} m}{c}, \frac{r^{2}}{c}\right)\).
Proof:
The given line is y = mx + c i.e., mx – y + c = 0 ……. (1)
The given circle is S = x² + y² = r² ……. (2)
Centre C = (0, 0), radius = r
Let P(x₁, y₁) be the point of contact.
Equation of the tangent is S₁ = 0, ⇒ x₁x + y₁y – r² = 0. ………. (3)

Equations (1) and (3) are representing the same line, therefore, \(\frac{x_{1}}{m}=\frac{y_{1}}{-1}=\frac{-r^{2}}{c}\) ⇒ x₁ = \(\frac{-r^{2} m}{c}\), y₁ = \(\frac{r^{2}}{c}\)
Point of contact is (x₁, y₁) = \(\left(-\frac{\mathrm{r}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{r}^{2}}{\mathrm{c}}\right)\)

Theorem: If P(x, y) is a point on the circle with centre C(α, β) and radius r, then x = α + r cos θ, y = β + r sin θ where 0 ≤ θ < 2π.

Note 1: The equations x = α + r cos θ, y = β + r sin θ, 0 ≤ θ < 2π are called parametric equations of the circle with centre (α, β) and radius r.

Note 2: A point on the circle x² + y² = r² is taken in the form (r cosθ, r sin θ). The point (r cosθ, r sin θ) is simply denoted as point θ.

Theorem: The equation of the chord joining two points θ₁ and θ₂ on the circle
x² + y² + 2gx + 2fy + c = 0 is (x + g)cos\(\frac{\theta_{1}+\theta_{2}}{2}\) + (y + f) sin \(\frac{\theta_{1}+\theta_{2}}{2}\) = r cos \(\frac{\theta_{1}+\theta_{2}}{2}\) where r is the radius of the circle.

Note 1: The equation of the chord joining the points θ₁ and θ₂ on the circle x² + y² = r² is x cos\(\frac{\theta_{1}+\theta_{2}}{2}\) + y sin\(\frac{\theta_{1}+\theta_{2}}{2}\) = r cos\(\frac{\theta_{1}-\theta_{2}}{2}\)

Note 2: The equation of the tangent at P(θ) on the circle (x + g) cos θ + (y + f) sin θ = \(\sqrt{g^{2}+f^{2}-c}\).

Note 3: The equation of the tangent at P(θ) on the circle x² + y² = r² is x cos θ + y sin θ = r.

Note 4: The equation of the normal at P(θ) on the circle x² + y² = r² is x sin θ – y cos θ = r.

Theorem:
If a line passing through a point P(x₁, y₁) intersects the circle S = 0 at the points A and B then PA.PB = |S₁₁|.

Corollary:
If the two lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 meet the coordinate axes in four distinct points then those points are concyclic ⇔ a₁a₂ = b₁b₂.

Corollary:
If the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 meet the coordinate axes in four distinct concyclic points then the equation of the circle passing through these concyclic points is (a₁x + b₁y + c₁) (a₂x + b₂y + c₂) – (a₁b₂ + a₂b₁)xy = 0.

Theorem:
Two tangents can be drawn to a circle from an external point.

Note:
If m₁, m₁ are the slopes of tangents drawn to the circle x² + y² = a² from an external point (x₁, y₁) then m₁ + m₂ = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\), m₁m₂ = \(\frac{y_{1}^{2}-a^{2}}{x_{1}^{2}-a^{2}}\).

Theorem:
If θ is the angle between the tangents through a point P to the circle S = 0 then tan = \(\frac{\theta}{2}=\frac{r}{\sqrt{S_{11}}}\) where r is the radius of the circle.
Proof:

Let the two tangents from P to the circle S = 0 touch the circle at Q, R and θ be the angle between
these two tangents. Let C be the centre of the circle. Now QC = r, PQ = \(\sqrt{S_{11}}\) and ∆PQC is a right angled triangle at Q.
∴ tan \(\frac{\theta}{2}=\frac{\mathrm{QC}}{\mathrm{PQ}}=\frac{\mathrm{r}}{\sqrt{\mathrm{S}_{11}}}\)

Theorem: The equation to the chord of contact of P(x₁, y₁) with respect to the circle S = 0 is S₁ = 0.

Theorem: The equation of the polar of the point P(x₁, y₁) with respect to the circle S = 0 is S₁ = 0.

Theorem: The pole of the line lx + my + n = 0 (n ≠ 0) with respect to x² + y² = a² is \(\left(-\frac{1 a^{2}}{n},-\frac{m a^{2}}{n}\right)\)
Proof :
Let P(x₁, y₁) be the pole of lx + my + n = 0 ……. (1)
The polar of P with respect to the circle is:
xx₁ + yy₁ – a² = 0
Now (1) and (2) represent the same line
∴ \(\frac{\mathrm{x}_{1}}{\ell}=\frac{\mathrm{y}_{1}}{\mathrm{~m}}=\frac{-\mathrm{a}^{2}}{\mathrm{n}}\) ⇒ x₁ = \(\frac{-\mathrm{la}^{2}}{\mathrm{n}}\), y₁ = \(\frac{-\mathrm{ma}^{2}}{\mathrm{n}}\)
∴ Pole P = \(\left(-\frac{1 a^{2}}{n},-\frac{m a^{2}}{n}\right)\)

Theorem: If the pole of the line lx + my + n = 0 with respect to the circle x² + y² + 2gx + 21y + c = 0 is (x₁, y₁) then \(\frac{x_{1}+g}{\ell}=\frac{y_{1}+f}{m}=\frac{r^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\) where r is the radius of the circle.
Proof:
Let P(x₁, y₁) be the pole of the line lx + my + n = 0 ……. (1)
The poiar of P with respect to S = 0 is S₁ = 0
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ (x₁ + g)x + (y₁ + f) + gx₁ + fy₁ + c = 0 …….. (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}=\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{gx} \mathrm{x}_{1}+\mathrm{gy} \mathrm{y}_{1}+\mathrm{c}}{\mathrm{n}}\) = k(say)
\(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}\) = k ⇒ x₁ + g = l k ⇒ x₁ = lk – g
\(\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}\) = k ⇒ y₁ + f = m k ⇒ y₁ = mk – f
\(\frac{g x_{1}+g y_{1}+c}{n}\) = k ⇒ gx₁ + gy₁ + c = nk
⇒ g(lk – g) + f(mk – f) + c = nk
⇒ k (lg + mf – n) = g² + f² – c = r² Where r is the radius of the circle ⇒ k = \(\frac{r^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\)
∴ \(\frac{\mathrm{x}_{1}+\mathrm{g}}{\ell}=\frac{\mathrm{y}_{1}+\mathrm{f}}{\mathrm{m}}=\frac{\mathrm{r}^{2}}{\lg +\mathrm{mf}-\mathrm{n}}\)

Theorem: The lines l₁x + m₁y + n₁ = 0 and l₂x + m₂y + n₁y = 0 are conjugate with respect to the circle x² + y₁ + 2gx + 2fy + c = 0 iffr₁ (l₁l₂ + m₁m₂) = (l₁g + m₁f – n₁) (l₂g + m₂f – n₂).

Theorem: The condition for the lines l₁x + m₁y + n₁ = 0 and l₂x + m₂y + n₂ = 0 to be conjugate with respect to the circle x² + y² = a² is a²(l₁l₂ + m₁m₂) = n₁n₂.

Theorem: The equation of the chord of the circle S = 0 having P(x₁, y₁) as its midpoint is S₁ = S₁₁.

Theorem: The length of the chord of the circle S = 0 having P(x₁, y₁) as its midpoint is 2\(\sqrt{\left|S_{11}\right|}\).

Theorem: The equation to the pair of tangents to the circle
S = 0 from P(x₁, y₁) is S²₁ = S₁₁S.
Proof:

Let the tangents from P to the circle S = 0 touch the circle at A and B.
Equation of AB is S₁ = 0.
i.e., x₁x + y₁y + g(x + x₁) + f(y + y₁) + c = 0 ———- (i)
Let Q(x₂, y₂) be any point on these tangents. Now locus of Q will be the equation of the pair of tangents drawn from P.
The line segment PQ is divided by the line AB in the ratio – S₁₁:S₂₂
⇒ \(\frac{P B}{Q B}=\left|\frac{S_{11}}{S_{22}}\right|\) ———— (ii)
But PB = \(\sqrt{S_{11}}\), QB = \(\sqrt{S_{22}}\) ⇒ \(\frac{P B}{Q B}=\frac{\sqrt{S_{11}}}{\sqrt{S_{22}}}\) ———— (iii)
From (ii) and (iii) ⇒ \(\frac{s_{11}^{2}}{s_{22}^{2}}=\frac{S_{11}}{S_{22}}\)
⇒ S₁₁S₂₂ = S²₁₂
Hence locus of Q(x₂, y₂) is S₁₁S = S²₁₂

Touching Circles: Two circles S = 0 and S’ = 0 are said to touch each other if they have a unique point P in common. The common point P is called point of contact of the circles S = 0 and S’ = 0.

Circle – Circle Properties: Let S = 0, S’ = 0 be two circle with centres C₁, C₂ and radii r₁, r₂ respectively.

• If C₁C₂ > r₁ + r₂ then each circle lies completely outside the other circle.
• If C₁C₂ = r₁ + r₂ then the two circles touch each other externally. The point of contact divides C₁C₂ in the ratio r₁ : r₂ internally.
• If |r₁ – r₂| < C₁C₂ < r₁ + r₂ then the two circles intersect at two points P and Q. The chord \(\overline{\mathrm{PQ}}\) is called common chord of the circles.
• If C₁C₂ = |r₁ – r₂| then the two circles touch each other internally. The point of contact divides C₁C₂ in the ratio r₁: r₂ externally.
• If C₁C₁ < |r₁ – r₂] then one circle lies completely inside the other circle.

Common Tangents: A line L = 0 is said to be a common tangent to the circle S = 0, S’ = 0 if L = 0 touches both the circles.

Definition: A common tangent L = 0 of the circles S = 0, S’= 0 is said to be a direct common tangent of the circles if the two circles S = 0, S’ = 0 lie on the same side of L = 0.

Centres of Similitude:
Let S = 0, S’ = 0 be two circles.

• The point of intersection of direct common tangents of S = 0, S’ = 0 is called external centre of similitude.
• The point of intersection of transverse common tangents of S = 0, S’ = 0 is called internal centre of similitude.

Theorem:
Let S = 0, S’ = 0 be two circles with centres C₁, C₂ and radii r₁, r₂ respectively. If A₁ and A₂ are respectively the internal and external centres of similitude circles s = 0, S’ = 0 then

• A₁ divides C₁C₂ in the ratio r₁ : r₂ internally.
• A₂ divides C₁C₂ in the ratio r₁ : r₂ internally.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 Random Variables and Probability Distributions to solve questions creatively.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Formulas

Random variable:
→ Suppose S is the sample space of a random experiment then any function X : S → R is called a random variable.

→ Let S be a sample space and X : S → R.be a random variable. The function F : R → R defined by F(x) = P(X ≤ x), is called probability distribution function of the random variable X.

→ A set ‘A’ is said to be countable if there exists a bijection from A onto a subset of N.

→ Let S be a sample space. A random variable X : S → R is said to be discrete or discontinuous if the range of X is countable.

→ If X : S → R is a discrete random variable with range {x₁, x₂, x₃, ……………. } then
\(\sum_{r=1}^{\infty}\) P(x = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁, x₂, x₃, …………….} If Σx_r P(X = x_r) exists, then Σx_r. P(X = x) is called the mean of the random variable X. It is denoted by µ or x . If Σ (x_r – µ)² P(X = X_r) exists, then Σ(x_r – µ)² P(X = X_r) is called variance of the random variable X. It is denoted by σ².

→ The positive square root of the variance is called the standard deviation of the Fandom variable x. It is denoted by σ.

Binomial distribution:
→ Let n be a positive integer and p be a real number-such that 0 ≤ p ≤ 1. A random variable x with range {0, 1, 2, 3, ……….. n} is said to follows (or have) binomial distribution or Bernoulli distribution with parameters n and p if P (X = r) = ⁿC_r p^r q^{n – r} for r = 0, 1, 2, ………. n where q = 1 – p. Its Mean µ = np and variance σ² = npq. n and p are called. parameters of the Binomial distribution.

Poisson distribution:
→ Let λ > 0 be a real, number. A, random variable x with range {0, 1, 2, ……… n} is said to follows (have) poisson distribution with parameter λ if P(X = r) = \(\frac{e^{-\lambda} \lambda^{r}}{r !}\) for r = 0, 1, 2, ……………. . Its Mean = λ and variance = λ. Its parameter is λ.

→ If X : S → R is a discrete random variable with range {x₁ x₂, x₃, …. } then \(\sum_{r=1}^{\infty}\) P (X = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁ x₂, x₃, …..} .If Σ x_r P(X = x_r) exists, then Σ x_r P(X = x_r) is called the mean of the random variable X. It is denoted by or x.

→ If Σ(x_r – μ)² P(X = x_r) exists, then Σ (x_r – μ)² P(X = x_r) is called variance of the random variable X. It is denoted by σ². The positive square root of the variance is called the standard deviation of the random variable X. It is denoted by σ

→ If the range of discrete random variable X is {x₁ x₂, x₃, …. x_n, ..} and P(X = x_n) = P_n for every Integer n is given then σ² + μ² = Σx_n²P_n

Binomial Distribution:
A random variable X which takes values 0, 1, 2, ., n is said to follow binomial distribution if its probability distribution function is given by
P(X = r) = ⁿc_rp^rq^n-r, r = 0,1,2, ……………. , n where p, q > 0 such that p + q = 1.

→ If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is p^r.

→ Mean and variance of the binomial distribution

• The mean of this distribution is \(\sum_{i=1}^{n}\) X_ip_i = latex]\sum_{X=1}^{n}[/latex] X. ⁿC_xq^n-xp^X = np,
• The variance of the Binomial distribution is σ² = npq and the standard deviation is σ = \(\sqrt{(n p q)}\).

→ The Poisson Distribution : Let X be a discrete random variable which can take on the values 0, 1, 2,… such that the probability function of X is given by
f(x) = P(X = x) = \(\frac{\lambda^{x} e^{-\lambda}}{x !}\), x = 0, 1, 2, ………….
where λ is a given positive constant. This distribution is called the Poisson distribution and a random variable having this distribution is said to be Poisson distributed.