## The locus of the middle points of chords of hyperbola 3x^2 – 2y^2 + 4x – 6y = 0 parallel to y = 2x is

Q: The locus of the middle points of chords of hyperbola 3x2 – 2y2 + 4x – 6y = 0 parallel to y = 2x is

(A) 3x – 4y = 4

(B) 3y – 4x + 4 = 0

(C) 4x – 4y = 3

(D) 3x – 4y = 2

Ans: (A)

Sol. Let mid point be (h, k).

Equation of a chord whose mid point is (h, k) would be T = S1

or 3xh – 2yk + 2( x+h) – 3( y+k) = 3h2 – 2k2 + 4h – 6k

⇒ x(3k + 2) – y ( 2k+3) – 2h + 3k – 3h2 + 2k2 = 0

it’s slope is $\large \frac{3h+2}{2k+3} = 2$(given)

⇒ 3h = 4k + 4

⇒ Required locus is 3x – 4y = 4.

Hence (A) is the correct answer.

## The angle between the hyperbola xy = c^2 and x^2 – y^2 = a^2 is

Q: The angle between the hyperbola xy = c2 and x2 – y2 = a2 is

(A) independent of c

(B) dependent on a

(C) always p/3

(D) none of these

Ans: (A) & (B)
Sol: Differentiating xy = c2 with respect to x, we get

$\large \frac{dy}{dx} = -\frac{y}{x}$

Differentiating x2 – y2 = a2 with respect to x, we get

$\large \frac{dy}{dx} = \frac{x}{y}$

Clearly the two curves always intersect orthogonally.

Hence (A) and (B) are the correct answers.

## The equation (x – α)^2 + (y – β)^2 = k(lx + my + n)^2 represents

Q: The equation (x – α)2 + (y – β)2 = k(lx + my + n)2 represents

(A) a parabola for k = (l2 + m2)-1

(B) an ellipse for 0 < k < (l2 + m2)-1

(C) a hyperbola for k > (l2 + m2)-1

(D) a point circle for k = 0.

Sol. (x – α)2 + (y – β)2 = k(lx + my + n)2

$\large = k(l^2 + m^2)(\frac{lx + my + n}{\sqrt{l^2 + m^2}})^2$

⇒  PS/PM = k( l2 + m2)

If k(l2 + m2) = 1 , ‘P’ lies on parabola

If k(l2 + m2) < 1 , ‘P’ lies on ellipse

If k(l2 + m2) >1 , ‘P’ lies on hyperbola

If k = 0, ‘P’ lies on a point circle.

Hence (A), (B), (C) and (D) are the correct answers.

## If a rectangular hyperbola (x – 1)(y – 2) = 4 cuts a circle x^2 + y^2 + 2gx + 2fy +c = 0 at points

Q: If a rectangular hyperbola (x – 1)(y – 2) = 4 cuts a circle x2 + y2 + 2gx + 2fy +c = 0 at points (3, 4), (5, 3), (2, 6) and (-1, 0), then the value of (g + f) is equal to

(A) 8

(B) -9

(C) -8

(D) 9

Sol. Let xiyi, i = 1, 2, 3, 4 be the points of intersection of hyperbola and circle, then

$\Large \frac{\Sigma x_i}{4} = \frac{-g+1}{2}$

g = -7/2

and $\Large \frac{\Sigma y_i}{4} = \frac{-f+1}{2}$

∴ f = -9/2

g + f = -8.

Hence (C) is the correct answer.

## The point of intersection of the curves whose parametric equations are x = t^2 + 1, y = 2t and x = 2s, y = 2/s, is given by

Q: The point of intersection of the curves whose parametric equations are x = t2 + 1, y = 2t and x = 2s, y = 2/s, is given by

(A) (1, -3)

(B) (2, 2)

(C) (-2, 4)

(D) (1, 2)

Ans: (B)
Sol: x = t2 + 1 , y = 2t

⇒ x-1 = y2/4

x = 2s , y = 2/s

⇒ xy = 4

For the point of intersection we have,

$\large \frac{4}{y}-1 = \frac{y^2}{4}$

⇒ y3 + 4y –16 = 0

⇒ y = 2 ⇒ x = 2

Hence (B) is the correct answer.

## The tangent at a point P on the hyperbola x^2/a^2 – y^2/b^2 = 1 meets one of its directrices in F.

Q: The tangent at a point P on the hyperbola $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ meets one of its directrices in F. If PF subtends an angle θ at the corresponding focus, then θ equals

(A) π/4

(B) π/2

(C) 3π/4

(D) π

Sol. Let directrix be x = a/e and focus be S(ae, 0). Let P( asecθ , btanθ) be any point on the curve,

Equation of tangent at ‘P’ is $\large \frac{x sec\theta}{a} – \frac{y tan\theta}{b} = 1$

Let ‘F’ be the intersection point of tangent of directrix, then

$\large F \equiv (a/e , \frac{b(sec\theta – e)}{e tan\theta})$

$\large m_{SF} = \frac{b(sec\theta – e)}{-a tan\theta (e^2 -1)}$

$\large m_{PS} = \frac{b tan\theta}{a(sec\theta -e)}$

⇒ mSF . mPS = –1

Hence (B) is the correct answer.

## The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y – 6 = 0,

Q: The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y – 6 = 0, the equation of the other asymptote is

(A) 4x – 3y + 17 = 0

(B) -4x – 3y + 17 = 0

(C) -4x + 3y + 1 = 0

(D) 4x + 3y + 17 = 0

Sol. We know that asymptotes of rectangular hyperbola are mutually perpendicular, thus other asymptote should be 4x + 3y + λ = 0

Intersection point of asymptotes is also the centre of the hyperbola.

Hence intersection point of 4x +3y + λ = 0 and 3x -4y -6 = 0 should lie on the line x- y -1= 0, using it λ can be easily obtained.

Hence (D) is the correct answer.

## For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

Q: For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

(A) $\frac{x^2}{4} – \frac{y^2}{12} = 1$

(B) $\frac{x^2}{12} – \frac{y^2}{4} = 1$

(C) $\frac{x^2}{16} – \frac{y^2}{4} = 1$

(D) $\frac{x^2}{4} – \frac{y^2}{16} = 1$

Sol. ae = 4, a = 2

⇒ e = 2

But b2 = a2(e2 – 1)

⇒ b2 = 12

Hyperbola is $\frac{x^2}{4} – \frac{y^2}{12} = 1$

Hence (A) is the correct answer.

## The point on the hyperbola x^2/24 – y^2/18 = 1 which is nearest to the line 3x + 2y+1 = 0 is

Q: The point on the hyperbola $\frac{x^2}{24} – \frac{y^2}{18} = 1$ which is nearest to the line 3x + 2y+1 = 0 is

(A) (6, 3)

(B) (–6, 3)

(C) (6, –3)

(D) (–6, –3)

Ans: (C)
Sol: Equation of tangent at $(\sqrt{24} sec\theta , \sqrt{18} tan\theta)$ is

$\large \frac{x sec\theta}{\sqrt{24}} – \frac{y tan\theta}{\sqrt{18}} =1$ ,

then point is nearest to the line 3x + 2y + 1 = 0,

so its slope = –3/2

$\large \frac{sec\theta}{\sqrt{24}} . \frac{\sqrt{18}}{tan\theta} = – \frac{3}{2}$

$\large sin\theta = -\frac{1}{\sqrt{3}}$

Hence the point is (6, –3).

Hence (C) is the correct answer.

## The line lx + my + n = 0 will be a normal to the hyperbola b^2x^2 – a^2y^2 = a^2b^2 if

Q: The line lx + my + n = 0 will be a normal to the hyperbola b2 x2 – a2 y2 = a2 b2 if

(A) $\large \frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2}$

(B) $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2}$

(C) $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n}$

(D) none of these

Sol. Equation of normal at (asecq, btanq) is;

ax cosq + by cotq = a2 +b2

Comparing it with lx+my + n = 0 we get

$\large \frac{a cos\theta}{l} = \frac{b cot\theta}{m} = \frac{a^2 + b^2 }{-n}$

$\large cos\theta = \frac{l(a^2 + b^2)}{-n a}$ and $\large cot\theta = \frac{m(a^2 + b^2)}{-n b}$

$\large sin\theta = \frac{b l}{a m}$

Thus , $\large \frac{b^2 l^2}{a^2 m^2} + \frac{l^2(a^2 + b^2)^2}{a^2 n^2} = 1$

or , $\large \frac{a^2}{l^2} – \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2}$

Hence (B) is the correct answer.