Let $\mathrm{e}$ be the eccentricity of the ellipse $\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1$. If…
Let $\mathrm{e}$ be the eccentricity of the ellipse $\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1$. If $\frac{1}{\mathrm{e}}$ is the eccentricity of a hyperbola, then the eccentricity of its conjugate hyperbola is
$\frac{4}{3}$
$\frac{3}{\sqrt{5}}$
$\frac{4}{\sqrt{5}}$
$\frac{3}{2}$
Solution
Given ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$
Since $b>a$ i.e. $9>4$. Hence
$
\begin{aligned}
& \Rightarrow e=\frac{1}{b} \sqrt{h^2-a^2}=\frac{1}{3} \sqrt{9-4} \\
& \Rightarrow e=\frac{\sqrt{5}}{3}
\end{aligned}
$
Let $e_H=\frac{1}{e}=\frac{3}{\sqrt{5}}$ be the eccentricity of hyperbola and $e_C$ be the eccentricity of the conjugate hypérbola.
We know that, for a hyperbola-
$
\begin{aligned}
& \Rightarrow \quad \frac{1}{e_H^2}+\frac{1}{e_C^2}=1 \Rightarrow\left(\frac{\sqrt{5}}{3}\right)^2+\frac{1}{e_C^2}=1 \\
& \Rightarrow e_C=3 / 2
\end{aligned}
$