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## AP Inter 2nd Year Maths 2B Question Paper May 2019

Time : 3 Hours

Max. Marks : 75

Note : This question paper consists of 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.

If x^{2} + y^{2} + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3) find g, f and its radius.

Solution:

Given circle equation is x^{2} + y^{2} + 2gx + 2fy – 12 = 0.

centre = (- g, – f)

⇒ (2,3) = (-g,-f)

⇒ – g = 2, – f = 3

⇒ g = -2, f = -3 and c = – 12

Radius = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{4+9-(-12)}\)

= \(\sqrt{4+9+12}\) = \(\sqrt{25}\) = 5.

Question 2.

Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x^{2} + y^{2} = 35.

Solution:

Let S ≡ x^{2} + y^{2} – 35 = 0

Since the points (1, 3) and (2, k) an conjugate with respect to the circle s = 0

∴ s_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} – 35 = 0

⇒ 1(2) + 3(k) – 35 = 0

⇒ 3k = 33 ⇒ k = 11.

Question 3.

Find the angle between the circles

x^{2} + y^{2} – 12x – 6y + 41 = 0

x^{2} + y^{2} + 4x + 6y – 59 = 0.

Solution:

Given circle equations are

x^{2} + y^{2} – 12x – 6y + 41 = 0 ……. (1)

x^{2} + y^{2} + 4x + 6y – 59 = 0 …… (2)

Question 4.

Find the co-ordinates of the points on the parabola y^{2} = 8x whose focal distance is 10.

Solution:

Given parabola equation is y^{2} = 8x

Here 4a = 8 ⇒ a = 2

∴ Focus = (a, 0) = (2, 0)

Let p(x_{1}, y_{1}) be a point on the parabola.

Given sp = 10

⇒ (x_{1} + 12) (x_{1} – 8) = 0

⇒ x_{1} = – 12 (or) x_{1} = 8

If x_{1} = -12 then \(y_1{ }^2\) = – 96 < 0

If x_{1} = 8 then \(y_1^2\) = 64 ⇒ y_{1} = ± 8

∴ Co-ordinates on the parabola are (8, 8) and (8, -8).

Question 5.

If the eccentricity of the hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola.

Solution:

If e_{1}, e_{2} are the eccentricities of hyperbola and conjugate hyperbola then \(\frac{1}{\mathrm{e}_1^2}\) + \(\frac{1}{\mathrm{e}_2^2}\) = 1

Question 6.

Evaluate : ∫\(\frac{d x}{1+e^x}\), x ∈ R

Solution:

Question 7.

Evaluate: ∫\(\frac{1}{\cos h x+\sinh x}\)dx on R.

Solution:

Question 8.

Evaluate \(\int_0^2|1-x| d x\)

Solution:

Question 9.

Evaluate \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x .\)

Solution:

Question 10.

Find the differential equation corresponding to xy = ae^{x} + be^{-x}, a and b are parameters.

Solution:

Give xy = ae^{x} + be^{-x}

diffefentiating w.r.to ‘x’ on both sides, we have

Section-B

II. Short Answer Type Questions.

- Attempt any five questions.
- Each question carries four marks.

Question 11.

Find the equation of tangent and normal at (3, 2) of the circle x^{2} + y^{2} – x – 3y – 4 = 0.

Solution:

Let S = x^{2} + y^{2} – x – 3y – 4 = 0.

The equation of tangent at (3, 2) of the circle s = 0 is s_{1} = 0

⇒ x(3) + y(2) – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0.

⇒ 3x + 2y – \(\frac{1}{2}\)x – \(\frac{3}{2}\) – \(\frac{3}{2}\)y – 3 – 4 = 0

⇒ \(\frac{5}{2}\)x + \(\frac{1}{2}\)y – \(\frac{17}{2}\) = 0

⇒ 5x + y – 17 = 0.

The equation of normal at (3, 2) of the circle s = 0 in

1(x – 3) – 5(y – 2) = 0

⇒ x – 3 – 5y + 10 = 0

⇒ x – 5y + 7 = 0.

Question 12.

Find the radical centre of the circles x^{2} + y^{2} – 4x – 6y + 5 = 0 x^{2} + y^{2} – 2x – 4y – 1 = 0, x^{2} + y^{2} – 6x – 2y = 0.

Solution:

Lets s ≡ x^{2} + y^{2} – 4x – 6y + 5 = 0

s^{1} = x^{2} + y^{2} – 2x – 4y – 1 = 0

s^{11} = x^{2} + y^{2} – 6x – 2y = 0

The radical axis of s = 0, s^{1} = 0 in s – s^{1} = 0

⇒ -2x – 2y + 6 = 0

⇒ x + y – 3 = 0 ……….. (1)

The radical axis of s = 0, s^{11} = 0 is s – s^{11} = 0

⇒ 2x – 4y + 5 = 0 ……… (2)

solving (1) and (2), we have

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 distance between foci 4\(\sqrt{2}\).

Solution:

Question 14.

Find the equation of the tangents to the ellipse 2x^{2} + y^{2} = 8 which are

(i) parallel to x – 2y – 4 = 0

(ii) perpendicular to x + y + 2 = 0.

Solution:

i) Parallel to x – 2y – 4 = 0

ii)

Perpendicular to x + y + 2 = 0

Slope of tangent be ‘1’ as it is perpendicular to above line

y = mx ± \(\sqrt{a^2 m^2+b^2}\)

y = x ± \(\sqrt{4+8}\)

y = x ± 2\(\sqrt{3}\).

⇒ x – y ± 2\(\sqrt{3}\) = 0.

Question 15.

Find the centre, eccentricity, foci, directrices and the length of the latus rectum of the hyperbola 4(y + 3)^{2} – 9(x – 2)^{2} = 1.

Solution:

Given hyperbola equation is 4(y + 3)^{2} – 9 (x – 2)^{2} = 1

Question 16.

Evaluate : \(\int_0^4\left(16-x^2\right)^{\frac{5}{2}}\)

Solution:

I = \(\int_0^4\left(16-x^2\right)^{\frac{5}{2}} d x\)

put x = 4 sin θ

dx = 4 cos θ dθ

Question 17.

Solve : \(\frac{d y}{d x}\) + y tan x = cos^{3} x.

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:

x^{2} + y^{2} + 2gx + 2fy + c_{1} = 0

Satisfies (2, 0), (0, 1) (4, 5) we get

4 + 0 + 4g + c_{1} = 0 ……. (i)

0 + 1 + 2g. 0 + 2f + c_{1} = 0 ……. (ii)

16 + 25 + 8g + 10f + c_{1} = 0 …….. (iii)

(ii) – (i) we get

– 3 – 4g + 2f = 0

4g – 2f = – 3 …….. (iv)

(ii) – (iii) we get

– 40 – 8g – 8f = 0 (or)

g + f = – 5 …… (v)

Solving(iv) and (v) we get

g = –\(\frac{13}{6}\) = –\(\frac{17}{6}\)

Substituting g and f values in equation (i) we get

4 + 4(-\(\frac{13}{6}\)) + c_{1} = 0

c_{1} = \(\frac{14}{3}\)

Now equation x^{2} + y^{2} – \(\frac{13}{3}\)x – \(\frac{17}{3}\)y + \(\frac{14}{3}\) = 0

Now circle passes through (0, c) then

c^{2} – \(\frac{17}{3}\)c + \(\frac{14}{3}\) = 0

3c^{2} – 17c + 14 = 0

⇒ (3c – 14) (c – 1) = 0

(or)

c = 1 or \(\frac{14}{3}\)

Question 19.

Show that the circles x^{2} + y^{2} – 6x – 2y + 1 = 0, x^{2} + y^{2} + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Solution:

Given circle equations are

S_{1} ≡ x^{2} + y^{2} – 6x – 2y + 1 = 0

S_{2} ≡ x^{2} + y^{2} + 2x – 8y + 13 = 0

Centre A = (3, 1), r_{1} = \(\sqrt{9+1-1}\) = 3

Centre B = (-1, 4), r_{2} = \(\sqrt{1+16-13}\) = 2

AB = \(\sqrt{(-1-3)^2+(4-1)^2}\)

= \(\sqrt{16+9}\) = 5

= r_{1} + r_{2}

AB = r_{1} + r_{2}

∴ The given circles touch each other externally.

Let ‘P’ be the common point of the circles S_{1} = 0, S_{2} = 0

P divides AB in the ratio 3:2 internally

∴ P = \(\left(\frac{3(-1)+2(3)}{3+2}, \frac{3(4)+2(1)}{3+2}\right)\)

= \(\left(\frac{3}{5}, \frac{14}{5}\right)\)

Common tangent equation is S_{1} – S_{2} = 0

x^{2} + y^{2} – 6x – 2y + 1 – (x^{2} + y^{2} + 2x – 8y + 13) = 0

⇒ x^{2} + y^{2} – 6x – 2y + 1 – x^{2} – y^{2} – 2x + 8y – 13 = 0

⇒ – 8x + 6y – 12 = 0

⇒ 4x – 3y + 6 = 0.

Question 20.

Derive the equation of a parabola in the standard form y^{2} = 4ax with diagram.

Solution:

Equation of a parabola in standard form :

To study the nature of the curve, we prefer its equation in the simplest possible form. We proceed as follows to derive such an equation.

Let S be the focus, l be the directrix as shown in fig. Let z be the projection of ‘S’ on l and ‘A’ be the midpoint of \(\overline{S Z}\). A lies on the parabola because SA = AZ. A is called the vertex of the parabola. Let \(\overline{Y A Y^{\prime}}\) be the straight line through A and parallel to the directrix. Now take \(\overline{\mathrm{zx}}\) as the X-axis and \(\overline{Y Y^{\prime}}\) as the Y- axis.

Then A is the origin (0, 0). Let S = (a, 0), (a>0). Then z = (-a, 0) and the equation of the directrix is x + a = 0.

If P (x, y) is a point on the parabola and PM is the perpendicular distance from P to the directrix l, then \(\frac{S P}{S M}\) = e = 1

∴ (SP)^{2} = (PM)^{2}

⇒ (x – a)^{2} + y^{2} = (x + a)^{2}

∴ y^{2} = 4ax

Conversely if P(x, y) is any point such that y^{2} = 4ax then

SP = \(\sqrt{(x-a)^2+y^2}\) = \(\sqrt{x^2+a^2-2 a x+4 a x}\) = \(\sqrt{(x+a)^2}\)

= |x + a| = PM

Hence P(x, y) is on the locus. In other words, a necessary and sufficient condition for the point P(x, y) to be on the parabola is that y^{2} = 4ax.

Thus the equation of the Parabola is y^{2} = 4ax.

Question 21.

Evaluate : ∫\(\sqrt{\frac{5-x}{x-2}}\) on (2, 5).

Solution:

I = \(\int \sqrt{\frac{5-x}{x-2}} d x\)

Question 22.

Obtain reduction formula I_{n} = ∫sin^{4} x.dx, n be a positive integer, n ≥ 2 and deduce the value of ∫sin^{4} x.dx.

Solution:

Question 23.

Find the area bounded between the curves y^{2} = 4ax, x^{2} = 4by (a > 0, b > 0).

Solution:

Given curve equations are y^{2} = 4ax, x^{2} = 4by (a > 0, b > 0).

⇒ \(\left(\frac{x^2}{4 b}\right)^2\) = 4ax

⇒ \(\frac{x^4}{166^2}\) = 4ax

⇒ x^{4} = 64 ab^{2}x

⇒ x^{4} – 64ab^{2}x = 0

⇒ x(x^{3} – 64ab^{2}) = 0

⇒ x = 0 (or) x = 4a^{1/3}b^{2/3}

Question 24.

Find the equation of a curve whose gradients is \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cos^{2}\(\frac{y}{x}\); where x > 0, y > 0 and which passes through the point

Solution: