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

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 :, $\frac{x^2}{9}-\frac{y^2}{16}=1$, $\frac{x^2}{16}-\frac{y^2}{9}=-1$, $\frac{x^2}{9}-\frac{y^2}{25}=1$, $\frac{x^2}{9}-\frac{y^2}{36}=1$

MARKS App · Accepted Answer

The eccentricity of $\frac{x^2}{16}+\frac{y^2}{25}=1$ is $\begin{aligned} e_1 & =\sqrt{1-\frac{16}{25}}=\frac{3}{5} \ \therefore e_2 & =\frac{5}{3}\left(\because e_1 e_2=1\right) \end{aligned}$ $\Rightarrow$ foci of ellipse $=(0, \pm 3)$ $\Rightarrow$ Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=-1$

Search any question & find its solution

Looking for more such questions to practice?