JEE Advanced · 2006 · 09 Apr 2006 · Hyperbola

If a hyperbola passes through the focus of the ellipse

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, and the product of eccentricities is 1, then

(A)the equation of hyperbola is $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ ✓ Correct
(B)the equation of hyperbola is $\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1$
(C)focus of hyperbola is $(5, 0)$ ✓ Correct
(D)focus of hyperbola is $(53, 0)$

Answer: (A), (C)

Solution

For the ellipse

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

For the hyperbola, let the equation be

\frac{x ^{2}}{a _{h}^{2}} - \frac{y ^{2}}{b _{h}^{2}} = 1

.

Given: product of eccentricities = 1, so

e_{2} = \frac{1}{e _{1}} = \frac{5}{3}

The hyperbola passes through the focus

(3, 0)

of the ellipse:

\frac{9}{a _{h}^{2}} - 0 = 1 \Rightarrow a_{h}^{2} = 9

For the hyperbola:

e_{2}^{2} = 1 + \frac{b _{h}^{2}}{a _{h}^{2}}

\frac{25}{9} = 1 + \frac{b _{h}^{2}}{9}

\frac{b _{h}^{2}}{9} = \frac{25}{9} - 1 = \frac{16}{9}

b_{h}^{2} = 16

So the equation is:

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

Foci of hyperbola:

c_{h} = a_{h}^{2} + b_{h}^{2} = 9 + 16 = 5

Foci are at

(\pm 5, 0)

JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options