If $e_{1}\$is the eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ and $e_{2}\$is the eccentricity of the hyperbola passing through the foci of the ellipse and$\ e_{1}e_{2} = 1$, then equation of the hyperbola is

Question

Dr. Abubacker · Accepted Answer

(b)
The eccentricity of $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ is $e_{1} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$
$	herefore\ e_{2} = \frac{5}{3}\ \ (\because e_{1}e_{2} = 1)$
and foci of given ellipse $(0, \pm 3)$
$	herefore 2b = 3 + 3 = 6\  \Rightarrow \ b = 3\  \Rightarrow \ \ b^{2} = 9$
$\Rightarrow \ \ a^{2} = 16$
$\Rightarrow$equation of hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = - 1$
Hence, (b) is the correct answer

Q. If $e_{1}\$is the eccentricity of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ and $e_{2}\$is the eccentricity of the hyperbola passing through the foci of the ellipse and$\ e_{1}e_{2} = 1$, then equation of the hyperbola is

Explanation

Reviewed by Dr. Abubacker

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