Inter 1st Year Maths Conic Sections Solutions Exercise 10c

Practicing the AP Board Solutions Class 11 Maths and Chapter 10 Inter 1st Year Maths Conic Sections Solutions Exercise 10c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Conic Sections Solutions Exercise 10c

Conic Sections Exercise 10c Solutions

Conic Sections Class 11 Exercise 10c Solutions – Conic Sections 10c Exercise Solutions

I.

Question 1.
Find the eccentricity of the ellipse \(\frac{x^2}{9}+\frac{y^2}{16}=1\).
Solution:
Eccentricity of the ellipse \(\frac{x^2}{9}+\frac{y^2}{16}=1\) (a < b) is \(\sqrt{\frac{b^2-a^2}{b^2}}\)
= \(\sqrt{\frac{16-9}{16}}\)
= \(\frac{\sqrt{7}}{4}\)

Question 2.
Find the length of the latus rectum of the ellipse x² + 16y² = 16.
Solution:
Given ellipse is x² + 16y² = 16
⇒ \(\frac{x^2}{16}+\frac{y^2}{1}\) = 1 (a > b)
Length of Latusrectum = \(\frac{2 b^2}{a}\)
= \(\frac{2 \times 1}{4}\)
= \(\frac {1}{2}\)

II. In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse.

Question 1.
\(\frac{x^2}{36}+\frac{y^2}{16}=1\)
Solution:
Given the ellipse equation:
\(\frac{x^2}{36}+\frac{y^2}{16}=1\)
This is in standard form: \(\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)\) = 1 with a² = 36, b² = 16
⇒ a = 6, b = 4
The major axis lies along the x-axis (since a² > b²).
The centre is at (0, 0)
⇒ c = \(\sqrt{\left(a^2-b^2\right)}\)
= \(\sqrt{(36-16)}\)
= 2√5
Results:
Vertices = (-6, 0) and (6, 0)
Foci = (-2√5, 0) and (2√5, 0)
Length of Major Axis = 2a = 12
Length of Minor Axis = 2b = 8
Eccentricity = e = \(\frac{c}{a}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}\) = 0.745
Length of Latusrectum = \(\frac{2 b^2}{a}=\frac{(2 \times 16)}{6}=\frac{32}{6}=\frac{16}{3}\) = 5.33

Question 2.
\(\frac{x^2}{4}+\frac{y^2}{25}=1\)
Solution:
\(\frac{x^2}{4}+\frac{y^2}{25}=1\)
This is in standard form
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
⇒ a² = 25, b² = 4
⇒ a = 5, b = 2
Vertices: Major axis is along the y-axis is (0, ±a) = (0, ±5)
Foci = c = \(\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}\)
So foci are (0, ±√21)
Length of Major axis = 2a = 2 × 5 = 10
Length of Minor axis = 2b = 2 × 2 = 4
Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{21}}{5}\) = 0.916
Length of latus rectum l = \(\frac{2 b^2}{a}=\frac{2 \times 4}{5}=\frac{8}{5}\) = 1.6

Question 3.
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Solution:
The given equation is \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
⇒ \(\frac{x^2}{4^2}+\frac{y^2}{3^2}=1\)
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 4 and b = 3, b < a
∴ \(\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}\)
Therefore, the coordinates of the foci are (±√7,0), and the vertices are (±4, 0).
Length of the major axis = 2a = 8.
Length of minor axis = 2b = 6.
Eccentricity, e = \(\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{7}}{4}\)
Length of latus rectum = \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 9}{4}=\frac{9}{2}\)

Question 4.
\(\frac{x^2}{25}+\frac{y^2}{100}=1\)
Solution:
The given equation is \(\frac{x^2}{25}+\frac{y^2}{100}=1\)
⇒ \(\frac{x^2}{5^2}+\frac{y^2}{10^2}=1\)
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 5 and b = 10, a < b.
∴ \(\sqrt{b^2-a^2}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3}\)
The coordinates of the foci are (0, ±5√3) and the vertices are (0, ±10)
Length of major axis = 2b = 20
Length of minor axis = 2a = 10
Eccentricity, e = \(\frac{\sqrt{b^2-a^2}}{b}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}\)
Length of latus rectum = \(\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 25}{10}\) = 5

Question 5.
\(\frac{x^2}{49}+\frac{y^2}{36}=1\)
Solution:
The given equaion is \(\frac{x^2}{49}+\frac{y^2}{36}=1\)
⇒ \(\frac{x^2}{7^2}+\frac{y^2}{6^2}=1\)
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 7, b = 6 and a > b.
∴ \(\sqrt{a^2-b^2}=\sqrt{49-36}=\sqrt{13}\)
The coordinates of the foci are (±√13, 0) and the vertices are (±7, 0).
Length of major axis = 2a = 14
Length of minor axis = 2b = 12
Eccentricity, e = \(\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{13}}{7}\)
Length of latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 36}{7}=\frac{72}{2}\)

Question 6.
\(\frac{x^2}{100}+\frac{y^2}{400}=1\)
Solution:
The given equation is \(\frac{x^2}{100}+\frac{y^2}{400}=1\)
⇒ \(\frac{x^2}{10^2}+\frac{y^2}{20^2}=1\)
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 10 and b = 20, a < b.
∴ \(\sqrt{b^2-a^2}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}\)
The coordinates of the foci are (0, ±10√3) and the vertices are (0, ±20)
Length of major axis = 2b = 40
Length of minor axis = 2a = 20
Eccentricity, e = \(\frac{\sqrt{b^2-a^2}}{b}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}\)
Length of latus rectum = \(\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 100}{20}\) = 10

Question 7.
36x² + 4y² = 144
Solution:
The given equation is 36x² + 4y² = 144
⇒ \(\frac{x^2}{4}+\frac{y^2}{36}=1\)
⇒ \(\frac{x^2}{2^2}+\frac{y^2}{6^2}=1\) ………(1)
On comparing equation (1) with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 2, b = 6 and a < b.
The coordinates of the foci are (0, 4√2) and the vertices are (0, ±6).
Length of major axis = 2b = 12
Length of minor axis = 2a = 4
Eccentricity, e = \(\frac{\sqrt{b^2-a^2}}{b}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}\)
Length of latus rectum = \(\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 4}{6}=\frac{4}{3}\)

Question 8.
16x² + y² = 16
Solution:
The given equation is 16x² + y² = 16
⇒ \(\frac{x^2}{1}+\frac{y^2}{16}=1\)
⇒ \(\frac{x^2}{1^2}+\frac{y^2}{4^2}=1\) …….(1)
On comparing equation (1) with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 1, b = 4 and a < b.
∴ \(\sqrt{b^2-a^2}=\sqrt{16-1}=\sqrt{15}\)
The coordinates of the foci are (0, ±√15), and the vertices are (0, ±4)
Length of major axis = 2b = 8
Length of minor axis = 2a = 2
Eccentricity, e = \(\frac{\sqrt{b^2-a^2}}{b}=\frac{\sqrt{15}}{4}\)
Length of latus rectum = \(\frac{2 a}{b}=\frac{2 \times 1}{4}=\frac{1}{2}\)

Question 9.
4x² + 9y² = 36
Solution:
The given equation is 4x² + 9y² = 36
⇒ \(\frac{x^2}{9}+\frac{y^2}{4}=1\)
⇒ \(\frac{x^2}{3^2}+\frac{y^2}{2^2}=1\) ………(1)
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),
we obtain a = 3, b = 2, and a > b
∴ \(\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}\)
The coordinates of the foci (±√5, 0) are, and the vertices are (±3, 0).
Length of major axis = 2a = 6
Length of the minor axis = 2b = 4
⇒ b = 2
Eccentricity, e = \(\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{5}}{3}\)
Length of latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 4}{3}=\frac{8}{3}\)

Question 10.
Find the equation of the ellipse with focus at (1, -1), e = \(\frac {2}{3}\) and directrix as x + y + 2 = 0.
Solution:
Focus S = (1, -1), e = \(\frac {2}{3}\), directrix is x + y + 2 = 0.
Let P(x, y) be any point on the ellipse.
∴ Equation of the ellipse is SP = e.PM
⇒ \(\sqrt{(x-1)^2+(y+1)^2}\) = \(\frac{2}{3}\left|\frac{\mathrm{x}+\mathrm{y}+2}{\sqrt{2}}\right|\)
⇒ 9[(x – 1)² + (y + 1)²] = 2(x + y + 2)²
⇒ 9[x² – 2x + 1 + y² + 2y + 1] = 2[x² + y² + 4 + 2xy + 4y + 4x]
⇒ 7x² – 4xy + 7y² – 26x + 10y + 10 = 0.

Question 11.
Find the equation of the ellipse in the standard form whose distance between foci is 2, and the length of the latus rectum is \(\frac {15}{2}\).
Solution:
Length latus recutm is \(\frac {15}{2}\)
⇒ \(\frac{2 b^2}{a}=\frac{15}{2}\)
⇒ 4b² = 15a
Distance between the foci is 2
⇒ 2ae = 2
⇒ ae = 1
4b² = 15a
⇒ 4a²(1 – e²) = 15a
⇒ 4a² – 4a²e² = 15a
⇒ 4a² – 15a – 4 = 0
⇒ (4a + 1) (a – 4) = 0
⇒ a = 4
4b² = 15a
⇒ 4b² = 15(4)
⇒ b² = 15
∴ Equation of the ellipse is \(\frac{x^2}{16}+\frac{y^2}{15}=1\)
⇒ 15x² + 16y² = 240

Question 12.
Find the equation of the ellipse in the standard form such that the distance between foci is 8 and the distance between directrices is 32.
Solution:
Distance between the foci = 8
⇒ 2ae = 8
⇒ ae = 4
Distance between the directrices = 32
⇒ \(\frac{2 \mathrm{a}}{\mathrm{e}}\) = 32
⇒ a = 16e
⇒ ae = 16e²
⇒ 4 = 16e²
⇒ e = \(\frac {1}{2}\)
ae = 4
⇒ a(\(\frac {1}{2}\)) = 4
⇒ a = 8
b² = a²(1 – e²)
= 64(1 – \(\frac {1}{4}\))
= 48
∴ Equation of the ellipse is \(\frac{x^2}{64}+\frac{y^2}{48}=1\)

In each of the following Exercises 13 to 23, find the equation for the ellipse that satisfies the given conditions:

Question 13.
Vertices (±5, 0), foci (±4, 0)
Solution:
It is given that b = 3, c = ae = 4, centre at the origin; foci on the x-axis.
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, b = 3, c = 4
It is known that a² = b² + c²
⇒ a² = 3² + 4²
⇒ a² = 25
⇒ a = 5
Thus, the equation of the ellipse is \(\frac{x^2}{5^2}+\frac{y^2}{3^2}=1\)
⇒ \(\frac{x^2}{25}+\frac{y^2}{9}=1\)

Question 14.
Vertices (0, ±13), foci (0, ±5)
Solution:
Vertices (0, ±13), foci (0, ±5)
Here, the vertices are on the y-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where b is the semi-major axis.
Accordingly, b = 13 and c = 5
It is known that b² = a² + c²
∴ 13² = a² + 5²
⇒ 169 = a² + 25
⇒ a² = 169 – 25
⇒ a = √144
⇒ a = 12
Thus, the equation of the ellipse is \(\frac{x^2}{12^2}+\frac{y^2}{13^2}=1\)
⇒ \(\frac{x^2}{144}+\frac{y^2}{169}=1\)

Question 15.
Vertices (±6, 0), foci (±4, 0)
Solution:
Vertices (±6, 0), foci (±4, 0).
Here, the vertices are on the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, a = 6, c = 4
It is known that a² = b² + c²
⇒ 6² = b² + 4²
⇒ 36 = b² + 16
⇒ b² = 26 – 16
⇒ b = √20
Thus, the equation of the ellipse is \(\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20})^2}=1\)
⇒ \(\frac{x^2}{36}+\frac{y^2}{20}=1\)

Question 16.
Ends of major axis (±3, 0), ends of minor axis (0, ±2).
Solution:
Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Here, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, a = 3 and b = 2.
Thus, the equation of the ellipse is \(\frac{x^2}{3^2}+\frac{y^2}{2^2}=1\)
⇒ \(\frac{x^2}{9}+\frac{y^2}{4}=1\)

Question 17.
Ends of major axis (0, ±√5), ends of minor axis (±1, 0).
Solution:
Ends of major axis (0, ±√5), ends of minor axis (±1, 0).
Here, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where b is the semi-major axis.
Accordingly, b = √5 and a = 1.
Thus, the equation of the ellipse is \(\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5})^2}=1\)
⇒ \(\frac{x^2}{1}+\frac{y^2}{5}=1\)

Question 18.
Length of major axis 26, foci (±5, 0)
Solution:
Length of major axis = 26
foci = (±5, 0)
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, 2a = 26
⇒ a = 13 and c = 5
It is known that a² = b² + c²
∴ 13² = b² + 5²
⇒ 169 = b² + 25
⇒ b² = 169 – 25
⇒ b = √144
⇒ b = 12
Thus, the equation of the ellipse is \(\frac{x^2}{13^2}+\frac{y^2}{12^2}=1\)
⇒ \(\frac{x^2}{169}+\frac{y^2}{144}=1\)

Question 19.
Length of minor axis 16, loci (0, ±6)
Solution:
Length of minor axis = 16
foci = (0, ±6)
Since the foci are on the y-axis, the major axis is along the y-axis.
Therefore, the equation of the ellipse will be the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where b is the semi-major axis.
Accordingly, 2a = 16
⇒ a = 8 and c = 6
It is known that b² = a² + c²
∴ b² = 8² + 6² = 64 + 36 = 100
⇒ b = √100 = 10
Thus, the equation of the ellipse is \(\frac{x^2}{8^2}+\frac{y^2}{10^2}=1\)
⇒ \(\frac{x^2}{64}+\frac{y^2}{100}=1\)

Question 20.
Foci (±3, 0), a = 4
Solution:
Foci (±3, 0), a = 4
Since the foci are on the x-axis, the major axis is along the x-axis.
Therefore, the equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a is the semi-major axis.
Accordingly, c = 3 and a = 4
It is known that a² = b² + c²
∴ 4² = b² + 3²
⇒ 16 = b² + 9
⇒ b² = 16 – 9
⇒ b² = 7
Thus the equation of the ellipse is \(\frac{x^2}{16}+\frac{y^2}{7}=1\)

Question 21.
b = 3, c = 4 centre at the origin; foci on the x axis.
Solution:
b = 3 (semi-minor axis)
c = 4 (distance from centre of focus)
Centre at origin (0, 0)
Foci lie on the x-axis
Major axis along x-axis c² = a² – b²
⇒ a² = c² + b²
⇒ a² = 25
⇒ a = 5
Equation of ellipse is \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
Foci on the axis = (±c, 0) = (±4, 0)

Question 22.
Centre at (0, 0), major axis on the y-axis, and passes through the points (3, 2) and (1, 6).
Solution:
Since the centre is at (0, 0) and the major axis is on the y-axis,
The equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) ……(1)
where b is the semi-major axis.
The ellipse passes through points (3, 2) and (1, 6)
⇒ \(\frac{9}{a^2}+\frac{4}{b^2}=1\) ………(2)
⇒ \(\frac{1}{a^2}+\frac{36}{b^2}=1\) ……..(3)
On solving equations (2) and (3),
we obtain a² = 10 and b² = 40
Thus, the equation of the ellipse is \(\frac{x^2}{10}+\frac{y^2}{40}=1\)
⇒ 4x² + y² = 40

Question 23.
The major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution:
Since the major axis is on the x-axis,
The equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) ……..(1)
where a is the semi-major axis.
The ellipse passes through points (4, 3) and (6, 2)
⇒ \(\frac{16}{a^2}+\frac{9}{b^2}=1\) ……(2)
⇒ \(\frac{36}{a^2}+\frac{4}{b^2}=1\) ………(3)
On solving equations (2) and (3),
we obtain a² = 52 and b² = 13
Thus, the equation of the ellipse is \(\frac{x^2}{52}+\frac{y^2}{13}=1\)
⇒ x² + 4y² = 52