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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.

- Attempt ALL questions.
- Each question carries TWO marks.

Question 1.

Obtain the parametric equations of the circle x^{2} + y^{2} – 6x + 4y – 12 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 6x + 4y – 12 = 0 ……. (1)

Comparing (1) with x^{2} + y^{2} + 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 x^{2} + y^{2} + 2ky = 0 is 1, then find k.

Solution:

The given equation of the circle is x^{2} + y^{2} + 2ky = 0.

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0

we get g = 0, f = k, c = 0

Let, the given point P(x_{1}, y_{1}) = (5, 4)

Given length of the tangent = 1

Question 3.

Show that the circles given by the following equations intersect each other orthogonally

x^{2} + y^{2} – 2x – 2y – 7 = 0, 3x^{2} + 3y^{2} – 8x + 29y = 0.

Solution:

Given equations of the circles are x^{2} + y^{2} – 2x – 2y – 7 = 0 ……. (1)

3x^{2} + 3y^{2} – 8x + 29y = 0

Comparing (1) with x^{2} + y^{2} + 2gx + 2fy + c = 0,

we get g = -1, f = -1, c = -7

Comparing (2) with

x^{2} + y^{2} + 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 y^{2} = 2x, whose focal distance is \(\frac{5}{2}\).

Solution:

Given equation of the parabola is y^{2} = 2x

Comparing with y^{2} = 4ax we get

4a = 2 ⇒ a = \(\frac{1}{2}\)

Let, P(x_{1}, y_{1}) be a point on the parabola y^{2} = 2x

Given that, focal distance = \(\frac{5}{2}\)

x_{1} + a = \(\frac{5}{2}\) ⇒ x_{1} + \(\frac{1}{2}\) = \(\frac{5}{2}\)

⇒ x_{1} = \(\frac{5}{2}\) – \(\frac{1}{2}\) ⇒ x_{1} = 2

Since P(x_{1}, y_{1}) lies on the parabola y^{2} = 2x then

\(y_1^2\) = 2x_{1} ⇒ \(y_1^2\) = 4 ⇒ y_{1} = ±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 : ∫e^{x} 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 = x^{2}, then X-axis and the lines x = -1, x = 2.

Solution:

Given y = x^{2}, 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

II. Short Answer type questions.

- Attempt ANY FIVE questions.
- Each question carries FOUR marks.

Question 11.

Find the pole of x + y + 2 = 0 with respect to the circle x^{2} + y^{2} – 4x + 6y -12 = 0.

Solution:

Given equation of the circle is x^{2} + y^{2} – 4x + 6y -12 = 0

Comparing this equation with x^{2} + y^{2} + 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. x^{2} + y^{2} + 2gx + 2fy + c = 0 is

Question 12.

Find the equation and length of the common chord of the following circles, x^{2} + y^{2} + 2x + 2y + 1 = 0, x^{2} + y^{2} + 4x + 3y + 2 = 0.

Solution:

Given equations of the circles are

S ≡ x^{2} + y^{2} + 2x + 2y + 1 = 0

S’ = x^{2} + y^{2} + 4x + 3y + 2 = 0

The equation of the common chord of the circles S = 0 and S’ = 0 is S – S’ = 0

(x^{2} + y^{2} + 2x + 2y + 1) – (x^{2} + y^{2} + 4x + 3y + 2) = .0

x^{2} + y^{2} + 2x + 2y + 1 – x^{2} – y^{2} – 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(x_{1}, y_{1}) be the point of contact.

The equation of the tangent at ‘P’ is S_{1} = 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

∴ lx_{1} + my_{1} + n = 0

l\(\left(\frac{-\mathrm{a}^2 l}{\mathrm{n}}\right)\) + m\(\left(\frac{-b^2 m}{n}\right)\) + n = 0

– a^{2}l^{2} – b^{2}m^{2} + n^{2} = 0 ⇒ a^{2}l^{2} + b^{2}m^{2} = n^{2}

Question 14.

Find the eccentricity and coordinates of the foci of the following ellipse. 3x^{2} + y^{2} – 6x – 2y – 5 = 0.

Solution:

Given equation of the ellipse is 3x^{2} + y^{2} – 6x – 2y – 5 = 0

(3x^{2} – 6x) + (y^{2} – 2y) – 5 = 0

3(x^{2} – 2x) + (y^{2} – 2y) – 5 = 0

3((x)^{2} – 2.1. x + 1^{2} -1^{2}) + ((y)^{2} – 2.1.y + 1^{2} – 1^{2}) – 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 3x^{2} – 4y^{2} = 12, which are

(i) parallel, and

(ii) perpendicular to the line y = x – 7.

Solution:

Given equation of the hyperbola is 3x^{2} – 4y^{2} = 12

\(\frac{x^2}{4}\) – \(\frac{y^2}{3}\) = 1

Here a^{2} = 4, b^{2} = 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 c^{2} = a^{2}m^{2} – b^{2}

k^{2} = 4(1)^{2} – 3

⇒ k^{2} = 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

c^{2} = a^{2}m^{2} – b^{2}

(-k)^{2} = 4 (-1)^{2} – 3

k^{2} = 4 – 3 ⇒ k^{2} = 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 + y^{2})dx = (tan^{-1}y – x)dy

Solution:

Section – C

III. Long Answer type questions.

- Attempt ANY FIVE questions.
- 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

x^{2} + y^{2} + 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. x^{2} + y^{2} – 2x – 6y + 6 = 0 and x^{2} + y^{2} = 1.

Solution:

∴ The two given circles each circle lies completely outside the other.

To find the external center of similitude (A_{2}):

Let, A_{2} be the external center of similitude. The externat center of similitude divides C_{1}C_{2} in the ratio r_{2} : r_{2} (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 (A_{1}):

Let, A_{1} be the internal center of similitude.

The internal center of similitude divides C_{1}C_{2} in the ratio r_{1} : r_{2}

(2 : 1) internally.

8y^{2} + 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 y^{2} = 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 I_{n} = ∫sin^{n}xdx

n being a positive integer, n ≥ 2 and deduce that value of ∫sin^{4} x dx.

Solution:

I_{n}= ∫sin^{n}xdx = ∫sin^{n-1}x . sin x dx

This is called Reduction formula for ∫sin^{n}xdx

If n is even, After successive Reduction we get

I_{0} = ∫sin^{0}xdx = x + c

If n is odd, After successive Reduction we get

I_{1} = ∫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: