# AP Inter 2nd Year Maths 2B Question Paper May 2015

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## AP Inter 2nd Year Maths 2B Question Paper May 2015

Time : 3 Hours
Max. Marks : 75

Note: This question paper contains three sections A, B and C.

Section – A
(10 × 2 = 20)

I. Very Short Answer type questions.

1. Attempt ALL questions.
2. Each question carries TWO marks.

Question 1.
Obtain the parametric equations of the circle x2 + y2 – 6x + 4y – 12 = 0.
Solution:
Given equation of the circle is x2 + y2 – 6x + 4y – 12 = 0 ……. (1)
Comparing (1) with x2 + y2 + 2gx + 2fy + c = 0
we get 2g = -6 ⇒ g = -3 ; 2f = 4 ⇒ f = 2 ; c = -12
Centre, C(h, k) = (-g, -f) = (3, -2)
Radius, r = $$\sqrt{g^2+f^2-c}$$
= $$\sqrt{(-3)^2+(2)^2+12}$$
= $$\sqrt{9+4+12}$$ = $$\sqrt{25}$$ = 5
∴ The parametric equations of the circle are
x = h + r cos θ = 3 + 5 cos θ
y = k + r sin θ = -2 + 5 sin θ, θ < 0 < 2π

Question 2.
If the length of the tangent from (5, 4) to the circle x2 + y2 + 2ky = 0 is 1, then find k.
Solution:
The given equation of the circle is x2 + y2 + 2ky = 0.
Comparing the given equation with x2 + y2 + 2gx + 2fy + c = 0
we get g = 0, f = k, c = 0
Let, the given point P(x1, y1) = (5, 4)
Given length of the tangent = 1

Question 3.
Show that the circles given by the following equations intersect each other orthogonally
x2 + y2 – 2x – 2y – 7 = 0, 3x2 + 3y2 – 8x + 29y = 0.
Solution:
Given equations of the circles are x2 + y2 – 2x – 2y – 7 = 0 ……. (1)
3x2 + 3y2 – 8x + 29y = 0
Comparing (1) with x2 + y2 + 2gx + 2fy + c = 0,
we get g = -1, f = -1, c = -7
Comparing (2) with
x2 + y2 + 2g’x + 2f’y + c’ = 0, we get
g’ = $$\frac{-4}{3}$$, f’ = $$\frac{29}{6}$$, c’ = 0
Now, 2gg’ + 2ff’ = 2(-1)$$\left(\frac{-4}{3}\right)$$ +2(-1)$$\left(\frac{29}{6}\right)$$
= $$\frac{8}{3}$$ – $$\frac{29}{3}$$ = $$\frac{-21}{3}$$ = -7
c + c’ = -7 + 0 = -7
∴ 2gg’ + 2ff’ = c + c’
∴ The two circles cut each other orthogonally.

Question 4.
Find the co-ordinates of the points on the parabola y2 = 2x, whose focal distance is $$\frac{5}{2}$$.
Solution:
Given equation of the parabola is y2 = 2x
Comparing with y2 = 4ax we get

4a = 2 ⇒ a = $$\frac{1}{2}$$
Let, P(x1, y1) be a point on the parabola y2 = 2x
Given that, focal distance = $$\frac{5}{2}$$
x1 + a = $$\frac{5}{2}$$ ⇒ x1 + $$\frac{1}{2}$$ = $$\frac{5}{2}$$
⇒ x1 = $$\frac{5}{2}$$ – $$\frac{1}{2}$$ ⇒ x1 = 2
Since P(x1, y1) lies on the parabola y2 = 2x then
$$y_1^2$$ = 2x1 ⇒ $$y_1^2$$ = 4 ⇒ y1 = ±2
∴ The required points are (2, 2), (2, -2).

Question 5.
If the eccentricity of hyperbola is $$\frac{5}{4}$$, then find the eccentricity of its conjugate hyperbola.
Solution:
The equation of the hyperbola is $$\frac{x^2}{a^2}$$ – $$\frac{y^2}{b^2}$$ = 1
Given that the eccentricity of the hyperbola e = $$\frac{5}{4}$$
The equation of the conjugate hyperbola is $$\frac{x^2}{a^2}$$ – $$\frac{y^2}{b^2}$$ = 1
The eccentricity of the conjugate hyperbola e’ = ?

Question 6.
Evaluate : $$\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^2}}$$dx on I = (0, 1).
Solution:

Question 7.
Evaluate : ∫ex cos x dx on R.
Solution:

Question 8.
Evaluate : $$\int_0^a \frac{d x}{x^2+a^2}$$
Solution:

Question 9.
Find the area bounded by the parabola y = x2, then X-axis and the lines x = -1, x = 2.
Solution:
Given y = x2, X – axis i.e., y = 0.
x = -1; x = 2

Question 10.
Find the order and degree of $$\left[\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\right]^{\frac{6}{5}}$$ = 6y.
Solution:

Section – B

1. Attempt ANY FIVE questions.
2. Each question carries FOUR marks.

Question 11.
Find the pole of x + y + 2 = 0 with respect to the circle x2 + y2 – 4x + 6y -12 = 0.
Solution:
Given equation of the circle is x2 + y2 – 4x + 6y -12 = 0
Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0,
we get g = – 2, f = 3, c = -12
Radius, r = $$\sqrt{g^2+f^2-c}$$ = $$\sqrt{4+9+12}$$ = $$\sqrt{25}$$ = 5
Given equation of the straight line is x + y + 2 = 0
Comparing this equation with lx + my + n = 0 we get l = 1, m = 1, n = 2
The pole of lx + my + n = 0 w.r. t. x2 + y2 + 2gx + 2fy + c = 0 is

Question 12.
Find the equation and length of the common chord of the following circles, x2 + y2 + 2x + 2y + 1 = 0, x2 + y2 + 4x + 3y + 2 = 0.
Solution:

Given equations of the circles are
S ≡ x2 + y2 + 2x + 2y + 1 = 0
S’ = x2 + y2 + 4x + 3y + 2 = 0
The equation of the common chord of the circles S = 0 and S’ = 0 is S – S’ = 0
(x2 + y2 + 2x + 2y + 1) – (x2 + y2 + 4x + 3y + 2) = .0
x2 + y2 + 2x + 2y + 1 – x2 – y2 – 4x – 3y – 2 = 0
2x + 2y + 1 – 4x – 3y – 2 = 0
2x – y -1 = 0
2x + y + 1 = 0
Centre of the circle S = 0 is C = (-1, -1)
Radius of the circle S = 0 is
r = $$\sqrt{(1)^2+(1)^2-1}$$ = $$\sqrt{1+1-1}$$ = $$\sqrt{1}$$ = 1
Now d = The perpendicular distance from the centre C (-1, -1) to the common chord 2x + y + 1 = 0.

Question 13.
Find the condition for the line lx + my + n = 0 to be a tangent to the ellipse $$\frac{x^2}{a^2}$$ + $$\frac{y^2}{b^2}$$ = 1
Solution:
Given equation of the ellipse is

$$\frac{x^2}{a^2}$$ + $$\frac{y^2}{b^2}$$ = 1
Suppose lx + my + n = 0 ……… (1)
is a tangent to the ellipse $$\frac{x^2}{a^2}$$ + $$\frac{y^2}{b^2}$$ = 1
Let P(x1, y1) be the point of contact.
The equation of the tangent at ‘P’ is S1 = 0
$$\frac{x x_1}{a^2}$$ + $$\frac{y_1}{b^2}$$ – 1 = 0 ……….. (2)
Now (1) & (2) represent the same line.

Since ‘P’ lies on the line lx + my + n = 0
∴ lx1 + my1 + n = 0
l$$\left(\frac{-\mathrm{a}^2 l}{\mathrm{n}}\right)$$ + m$$\left(\frac{-b^2 m}{n}\right)$$ + n = 0
– a2l2 – b2m2 + n2 = 0 ⇒ a2l2 + b2m2 = n2

Question 14.
Find the eccentricity and coordinates of the foci of the following ellipse. 3x2 + y2 – 6x – 2y – 5 = 0.
Solution:
Given equation of the ellipse is 3x2 + y2 – 6x – 2y – 5 = 0
(3x2 – 6x) + (y2 – 2y) – 5 = 0
3(x2 – 2x) + (y2 – 2y) – 5 = 0
3((x)2 – 2.1. x + 12 -12) + ((y)2 – 2.1.y + 12 – 12) – 5 = 0
3((x – 1)2 – 1) + ((y – 1)2 – 1) – 5 = 0
3(x – 1)2 – 3 + (y – 1)2 – 1 – 5 = 0
3(x – 1)2 + (y – 1)2 – 9 = 0
3(x – 1)2 + (y – 1)2 = 9

Question 15.
Find the equation of the tangents to the hyperbola 3x2 – 4y2 = 12, which are
(i) parallel, and
(ii) perpendicular to the line y = x – 7.
Solution:
Given equation of the hyperbola is 3x2 – 4y2 = 12
$$\frac{x^2}{4}$$ – $$\frac{y^2}{3}$$ = 1
Here a2 = 4, b2 = 3
Given equation of the straight line is y = x – 7 ⇒ x – y – 7 = 0

i) The equation of the tangent parallel to the line x – y – 7 = 0 is
x – y + k = 0 ………………. (i)
y = x + k
Comparing with y = mx + c, we get m = 1, c = k
Since equation (1) is a tangent to the given hyperbola then c2 = a2m2 – b2
k2 = 4(1)2 – 3
⇒ k2 = 1
⇒ k = ±1
Substitute the value of k in equation (1)
∴ The required parallel tangents are x – y ± 1 = 0

ii) The equation of the tangent perpendicular to the line x – y – 7 = 0 is x + y + k = 0 ……… (2)
⇒ y = -x- k
Comparing with y = mx + c, we get m = -1, c = – k
Since equation (2) is a tangent to the given hyperbola then
c2 = a2m2 – b2
(-k)2 = 4 (-1)2 – 3
k2 = 4 – 3 ⇒ k2 = 1 ⇒ k = ±1
Substitute the value of k in equation (2)
The required perpendicular tangents are x + y ± 1 = 0.

Question 16.
Evaluate : $$\int_0^{\pi / 2} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x} d x$$
Solution:

Question 17.
Solve :(1 + y2)dx = (tan-1y – x)dy
Solution:

Section – C

1. Attempt ANY FIVE questions.
2. Each question carries SEVEN marks.

Question 18.
If (2, 0), (0,1), (4, 5) and (0, c) are concyclic, then find c ?
Solution:
Let the equation of the required circle is
x2 + y2 + 2gx + 2fy + c = 0 …….. (1)
Since eq. (1) passes through the point (2, 0) then (2)2 + (0)2 + 2g(2) + 2f(0) + c = 0
4 + 4g + c = 0 ⇒ 4g + c = -4 ………… (2)
Since eq. (1) passes through the point (0,1) then
(0)2 + (-1)2 + 2g(0) + 2f(1) + c = 0 ⇒ 2f + c = -1 ………. (3)
Since eq. (1) passes through the point (4, 5) then
(4)2 + (5)2 + 2g(4) + 2f(5) + c = 0
16 + 25 + 8g + 10f + c = 0 ⇒ 8g + 10f + c = – 41 ………. (4)
From (2) & (3)

Question 19.
Find the equation to all possible common tangents of the circles. x2 + y2 – 2x – 6y + 6 = 0 and x2 + y2 = 1.
Solution:

∴ The two given circles each circle lies completely outside the other.
To find the external center of similitude (A2):
Let, A2 be the external center of similitude. The externat center of similitude divides C1C2 in the ratio r2 : r2 (2 : 1) externally.

Comparing coefficient of x on both sides 81 + k = -2 ……… (3)
Comparing coefficient of ‘y’ on both sides
-6l = -6 ⇒ l = 1
From(3), 8 + k = -2; k = -10
∴ The equations to the direct common tangents are
x + 1 = 0 and 8x – 6y – 10 = 0 ⇒ 4x – 3y – 5 = 0
To find the internal center of similitude (A1):
Let, A1 be the internal center of similitude.
The internal center of similitude divides C1C2 in the ratio r1 : r2
(2 : 1) internally.

8y2 + 6xy – 6x – 18y + 10 = (y + l) (6x + 8y + k)
Comparing coefficient of x on both sides 6l= -6 ⇒ l = -1
Comparing coefficient of y on both sides
8l + k = -18 ⇒ -8 + k = -18 ⇒ k = -10
∴ The equations to the transverse common tangents are
y – 1 = 0 and 6x + 8y – 10 = 0 ⇒ 3x + 4y – 5 = 0
∴ The equations of all common tangents are
x + 1 = 0, 4x – 3y – 5 = 0, y – 1 = 0, 3x + 4y – 5 = 0.

Question 20.
Derive equation of a Parabola in a standard form.
Solution:
The equation of a parabola in the standard form is y2 = 4ax.
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 (Foot of the perpendiculars) of P, S on the directrix, L = 0 respectively.
Let, ‘N’ be the projection of P and SZ.
Let, ‘A’ be the mid point of SZ.
Since, SA = AZ, ‘A’ lies on the parabola. .
Let, AS = a.
Take 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 = (x, y)
Now, PM = NZ = AN + AZ = x + a
P lies on the parabola

Question 21.
Obtain reduction formula In = ∫sinnxdx
n being a positive integer, n ≥ 2 and deduce that value of ∫sin4 x dx.
Solution:
In= ∫sinnxdx = ∫sinn-1x . sin x dx

This is called Reduction formula for ∫sinnxdx
If n is even, After successive Reduction we get
I0 = ∫sin0xdx = x + c
If n is odd, After successive Reduction we get
I1 = ∫sinx dx = -cosx + c

Question 22.
Find ∫$$\frac{d x}{3 \cos x+4 \sin x+6}$$
Solution:

Question 23.
Evaluate $$\int_0^\pi \frac{x \sin x}{1+\sin x} d x$$
Solution:

Question 24.
Solve : $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x-y+3}{2 x-2 y+5}$$
Solution: