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## AP Inter 2nd Year Maths 2B Question Paper May 2017

Time : 3 Hours

Max. Marks : 75

Note:

This question paper consists of THREE sections A, B and C.

Section – A

I. Very Short Answer Type Questions.

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

Question 1.

If x^{2} + y^{2} + 2gx + 2fy = 0 represents a circle with centre (-4, -3), then find g, f and the radius of the circle.

Solution:

Centre = (-g, -f) = (-4, -3) given

⇒ g = 4 and f = 3.

Radius of the circle = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{16+9}\) = 5

Question 2.

If the length of the tangent from (2, 5) to the circle x^{2} + y^{2} – 5x + 4y + k = 0 is \(\sqrt{37}\), then find the value of k.

Solution:

Given P(2, 5) = (x_{1}, y_{1}) and S = x^{2} + y^{2} – 5x + 4y + k = 0.

Question 3.

Find the angle between the circles x^{2} + y^{2} + 4x – 14y + 28 = 0, x^{2} + y^{2} + 4x – 5 = 0.

Solution:

x^{2} + y^{2} + 4x – 14y + 28 = 0 ……. (1)

x^{2} + y^{2} + 4x – 5 = 0 ………. (2)

Here g – 2, f = -7 : c = 28; g’ = 2; f = 0, c’ = -5.

Let θ be the angle between the circles (1) and (2)

cos θ = \(\frac{28-5-2(2)(2)-2(-7)(0)}{2 \sqrt{4+49-28} \sqrt{4+0+5}}\) = \(\frac{1}{2}\)

∴ θ = 60°

Question 4.

Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).

Solution:

The abscissae of the vertex and focus are equal to S. Hence the axis of the parabola is x = 3 a line parallel to y – axis, focus is above the vertex.

a = distance between focus and vertex = 3.

∴ Equation of the parabola (x – 3)^{2} = 4(3) (y + 2)

i.e., (x – 3)^{2} = 12 (y + 2).

Question 5.

If the angle between the asymtotes is 30°, then find the eccentricity of the hyperbola.

Solution:

Angle between a symptotes of hyperbola

Question 6.

Evaluate ∫sec^{2}x ∙ cosec^{2}x dx

Solution:

Question 7.

Evaluate \(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x\)

Solution:

Question 8.

Evaluate \(\int_0^4|2-x| d x\)

Solution:

Question 9.

Find the value of \(\int_0^{2 \pi}\)sin^{4} x ∙ cos^{6} x dx

Solution:

Let f(x) = sin^{4}x ∙ cos^{6} x dx

sin f(2π – x) = f(π – x) = f(x)

Question 10.

Form the differential equation of the curve y = a cos (nx + b) (where a, b are parameters).

Solution:

y = a cos(nx + b); (a, b)

Given equation is y = a cos (nx + b)

∴ y_{1} = -an sin(nx + b)

= -an^{2} cos (nx + b)

= -n^{2}y

∴ y_{1} + n^{2}y = 0 ⇒ \(\frac{d^2 y}{d x^2}\) + n^{2}y = 0 is the required differential equation obtained on elimination of a and b.

Section – B

(5 × 4 = 20)

II. Short Answer Type Questions.

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

Question 11.

Show that x + y + 1 = 0 touches the circle

x^{2} + y^{2} – 3x + 7y + 14 = 0 and find its point of contact.

Solution:

The equation of the given circle is

x^{2} + y^{2} – 3x + 7y + 14 = 0 ……… (1)

If the line x + y + 1 = 0 ………. (2)

touches the circle (1)

Then the perpendicular distance from the centre of the circle

C\(\left(\frac{3}{2},-\frac{7}{2}\right)\) to the line circle.

Hence the circle (1) touches the line (2).

Let P(h, k) be the point of contact of circle (1) and line (2), the P(h, k) will be the foot of the perpendicular drawn from C\(\left(\frac{3}{2},-\frac{7}{2}\right)\) to the line x + y + 1 = 0

∴ The point of contact of the given line with the circle (1) is (2, -3).

Question 12.

If x + y = 3 is the equation of the chord AB of the circle

x^{2} + y^{2} – 2x + 4y – 8 = 0. Find the equation of the circle having AB as diameter.

Solution:

Let S = x^{2} + y^{2} – 2x + 4y – 8 = 0 be the given circle.

Let L = x + y – 3 = 0 be the given line.

Given that L = 0 intersects the circle S = 0 at A, B..

Equation of the circle passing through the point of intersection of S = 0 and L = 0 is S + λL = 0 where λ is a constant.

(x^{2} + y^{2} – 2x + 4y – 8) + λ(x + y – 3) = 0

⇒ x^{2} + y^{2} + (λ – 2)x + (λ + 4)y – (8 + 3λ) = 0 ……. (1)

Centre of the circle = \(\left[-\left(\frac{\lambda-2}{2}\right)-\left(\frac{\lambda+4}{2}\right)\right]\)

Given that the line L = 0 is a diameter of the circle (1)

∴ The equation of the required circle from (1) is

x^{2} + y^{2} + (- 4 – 2)x + (- 4 + 4)y – 8 – 3(- 4) = 0

⇒ x^{2} + y^{2} – 6x + 4 = 0

Question 13.

Find the equation of the ellipse in the standard form such that distance between foci is 8 and distance between directrices is 32.

Solution:

Given that the distance between foci = 8.

∴ 2ae = 8 ⇒ ae = 4 ……. (1)

The distance betwen the directrices is ZZ’ = 32

⇒ 2\(\frac{\mathrm{a}}{\mathrm{e}}\) = 32 ⇒ \(\frac{a}{e}\) = 16 ………. (2)

From (1) and (2), (ae)\(\left(\frac{a}{e}\right)\) = (4) (16)

⇒ a^{2} = 64 ⇒ a = 8

b^{2} = a^{2} (1 – e^{2}) = a^{2} – a^{2}e^{2} = 64 – 16 = 48

∴ The equation of the required ellipse is

\(\frac{x^2}{a^2}\) + \(\frac{\mathrm{y}^2}{\mathrm{~b}^2}\) = 1 ⇒ \(\frac{x^2}{64}\) + \(\frac{y^2}{48}\) = 1.

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 and

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

Solution:

Given equation of ellipse is 2x^{2} + y^{2} = 8

⇒ \(\frac{x^2}{4}\) + \(\frac{y^2}{8}\) = 1

Let S = \(\frac{x^2}{4}\) + \(\frac{y^2}{8}\) – 1 = 0 ……… (1)

and compare with general equation, then we get a^{2} = 4 and b^{2} = 8

⇒ a = 2 ; b = 2\(\sqrt{2}\)

i) Parallel to x – 2y – 4 = 0

Given lines is x – 2y – 4 = 0 ………… (2)

Equation of any line parallel to x – 2y – 4 = 0 is

x – 2y + k = 0 ……. (3)

⇒ 2y = x + k ⇒ y = \(\frac{x}{2}\) + \(\frac{k}{2}\)

Where m = \(\frac{1}{2}\) and c = \(\frac{k}{2}\)

If (3) is a tangent to (1) them

c^{2} = a^{2}m^{2} + b^{2}

⇒ \(\frac{k^2}{4}\) = 4\(\left(\frac{1}{4}\right)\) + 8 ⇒ \(\frac{k^2}{4}\) = 1 + 8

⇒ k^{2} = 36

⇒ k = ±6

∴ The equation of the required tangent from (3) is x – 2y ± 6 = 0

ii) Perpendicular to x + y – 2 = 0

Given line is x + y – 2 = 0 …. (4)

Equation of any line perpendicular to (4) is

x – y + k = 0 ……. (5)

∴ y = x + k and m = 1; c = k

∴ By the condition for tangency

c^{2} = a^{2}m^{2} + b^{2}

⇒ k^{2} = 4(1) + 8

⇒ k^{2} = 12 ⇒ k = ±2\(\sqrt{3}\)

∴ Equation of the required line from (5) is x – y ± 2\(\sqrt{3}\) = 0.

Question 15.

Find the centre, foci, eccentricity and length of the latus rectum of the hyperbola 16y^{2} – 9x^{2} = 144.

Solution:

The given equation of hyperbola is 16y^{2} – 9x^{2} = 144

Question 16.

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

Solution:

Question 17.

Solve the differential equation \(\frac{d y}{d x}\) + \(\frac{4 x}{1+x^2}\)Y = \(\frac{1}{\left(1+x^2\right)^2}\)

Solution:

Section – C

III. Long Answer Type Questions.

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

Question 18.

Show that the four points (1, 1), (-6, 0), (-2, 2), (-2, -8) are concyclic and find the equation of the circle on which they lie.

Solution:

Given points (1, 1), (-6, 0), (-2, 2) and (-2, – 8).

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……… (1)

Since, (1) passes through the point (1, 1)

1^{2} + 1^{2} + 2g(1) + 2f(1) + c = 0

2 + 2g + 2f + c = 0

2g + 2f + c = – 2 ……… (2)

Since, (1) passes through the point (-6,0)

(-6)^{2} + 0^{2} + 2g(-6) + 0 + c = 0

36 + c – 12g = 0

– 12g + c = -36 …….. (3)

Since, (1) passes through the point (-2, 2)

(-2)^{2} + (2)^{2} + 2g(-2) + 2f(2) + c = 0

4 + 4 – 4g + 4f + c = 0

-4g + 4f + c = -8 ……… (4)

From (2) and (3),

Now, substitute the values of g, f, c in (1)

∴ The equation of the required circle is

x^{2} + y^{2} + 2(2) x + 2(3)y – 12 = 0

x^{2} + y^{2} + 4x + 6y – 12 = 0 ………. (7)

Now, substituting the point (-2, -8) in (1)

(-2)^{2} + (-8)^{2} + 2(2)(-2) + 2(3)(-8) – 12 = 0

4 + 64 – 8 – 48 – 12 = 0

68 – 68 = 0 ⇒ 0 = 0

∴ Given points are concyclic.

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

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:

Let S ≡ x^{2} + y^{2} – 6x – 2y + 1 = 0 and

S’ ≡ x^{2} + y^{2} + 2x – 8y + 13 = 0 be the given circles.

Then centre of circles are C_{1} = (3, 1) and C_{2} = (-1, 4)

Radius of circles are r_{1} = \(\sqrt{9+1-1}\) = 3

and r_{2} = \(\sqrt{1+16-13}\) = 2

Distance between centres

C_{1}C_{2} = \(\sqrt{(3+1)^2+(1-4)^2}\) = \(\sqrt{16+9}\) = 5 and r_{1} + r_{2} = 3 + 2 = 5

Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touch externally at a point. Let P be the point of contact of circles such that r_{1} : r_{2} = 3 : 2.

Since P divides C_{1}, C_{2} internally in the ratio 3 : 2 coordinates of point of contact.

The equation of common tangent is S – S’ = 0

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

⇒ – 8x + 6y – 12 = 0

⇒ 4x – 3y + 6 = 0

∴ The equation of the common tangent at the point of contact is 4x – 3y + 6 = 6

Question 20.

Show that the equation of common tangents to the circle x^{2} + y^{2} = 2a^{2} and the parabola y^{2} = 8ax are y = ±(x + 2a).

Solution:

Given parabola is y^{2} = 8ax

Equation of any tangent to the given parabola having slope ‘m’ is

y = mx + \(\frac{2 a}{m}\) ⇒ m^{2}x – my + 2a = 0 …… (1)

This is also a tangent to the circle x^{2} + y^{2} = 2a^{2}

So r = d. Here r = \(\sqrt{2}\)a and

d = Perpendicular distance .from centre (0, 0) to line (1)

Question 21.

Evaluate \(\int \frac{2 x+5}{\sqrt{x^2-2 x+10}} d x\)

Solution:

Question 22.

Obtain the reduction formula for I_{n} = ∫sec^{n} x dx, n is a positive integer, n ≥ 2 and deduce the value of ∫sec^{5} x dx.

Solution:

Question 23.

Show that the area of the region bounded by \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (ellipse) is πab. Also deduce the area of the circle x^{2} + y^{2} = a^{2}.

Solution:

Given equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.

Since the ellipse is symmetric with respect to coordinate axes we have

∴ Area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 is π ab sq. units.

Substituting b = a in the above result we have the area of circle given by

π a (a) = π a^{2} sq. units.

Question 24.

Solve the differential equation (x^{2} – y^{2}) dx – xy dy = 0.

Solution:

(x^{2} – y^{2})dx – xy dy = 0

\(\frac{d y}{d x}\) = \(\frac{x^2-y^2}{x y}\) ……… (1)