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

• A. $\frac{x^2}{9}-\frac{y^2}{16}=1$
• B. $\frac{x^2}{16}-\frac{y^2}{9}=-1$
• C. $\frac{x^2}{9}-\frac{y^2}{25}=1$
• D. $\frac{x^2}{25}-\frac{y^2}{9}=1$

Answer 1

The correct option is B $\frac{x^2}{16}-\frac{y^2}{9}=-1$

We have $e_1=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
$\because e_1{e}_2=1\Rightarrow e_2=\frac{5}{3}$
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of focii of the ellipse are $(0,\pm b{e}_1)$ or $(0,\pm 3)$.
Let hyperbola is, $\frac{y^2}{b^2}-\frac{x^2}{a^2}=\mathrm{1..}\left(1\right)$
given hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, required hyperbola is, $\frac{x^2}{16}-\frac{y^2}{9}=-1$
Hence, option 'A' is correct.