Question

If $e$ is eccentricity of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ and $e^{\prime }$ is eccentricity of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b)$ then

• A. $e=e^{\prime }$
• B. $e{e}^{\prime }=1$
• C. $\frac{1}{e^2}+{\frac{1}{\left(e^{\prime }\right)}}^2=1$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{1}{e^2}+{\frac{1}{\left(e^{\prime }\right)}}^2=1$

We have,

$\begin{array}{c}e=\sqrt{1-\frac{b^2}{a^2}}\left(b<a\right)\\ e^{{}^{\prime }}=\sqrt{1-\frac{a^2}{b^2}}\left(a>b\right)\end{array}$

Putting the value of $e$ and $e^{\prime }$ in equation.

$\begin{array}{c}\Rightarrow \frac{1}{e^2}+\frac{1}{{\left(e^{\prime }\right)}^2}=1\\ \Rightarrow \frac{1}{1-\frac{b^2}{a^2}}+\frac{1}{1-\frac{a^2}{b^2}}=1\\ \Rightarrow \frac{a^2}{a^2-b^2}-\frac{b^2}{a^2-b^2}=1\end{array}$

$L.H.S=R.H.S$

Hence, proved

Option $C$ is correct answer.