## 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.