Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{\mathrm{x}^2}{\mathrm{~b}^2}+\frac{\mathrm{y}^2}{25}=1 \quad$ and the hyperbola $\quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$, respectively. If $\mathrm{b} \lt 5$ and $\mathrm{e}_1 \mathrm{e}_2=1$, then the…

Question

Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{\mathrm{x}^2}{\mathrm{~b}^2}+\frac{\mathrm{y}^2}{25}=1 \quad$ and the hyperbola $\quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$, respectively. If $\mathrm{b} \lt 5$ and $\mathrm{e}_1 \mathrm{e}_2=1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{e}_1^2=1-\frac{\mathrm{b}^2}{25} \quad \mathrm{e}_2^2=1-\frac{\mathrm{b}^2}{16} \\ & \therefore \mathrm{e}_1^2 \mathrm{e}_2^2=1 \\ & \left(1-\frac{\mathrm{b}^2}{25}\right)\left(1+\frac{\mathrm{b}^2}{16}\right)=1 \\ & \Rightarrow 2+\frac{\mathrm{b}^2}{16}-\frac{\mathrm{b}^2}{25}-\frac{\mathrm{b}^2}{400}=1 \\ & \Rightarrow \frac{9 \mathrm{~b}^2}{400}=\frac{\mathrm{b}^4}{400} \\ & \mathrm{~b}^2=9 \\ & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=0 \\ & \mathrm{e}_1 \sqrt{1-\frac{9}{25}} \\ & \mathrm{e}_1=\frac{4}{5}
\end{aligned}$
Focii : - $(0, \pm 4) \quad( \pm 5,0)$
ellipse passing through all four foci
$\begin{aligned}
& \frac{x^2}{25}+\frac{y^2}{16}=1 \\ & e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}
\end{aligned}$

Let $e_1$ and $e_2$ be the eccentricities of the ellipse…

Solution