Question

$E_1=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0,(a>b)$ and $E_2=\frac{x^2}{k^2}+\frac{y^2}{b^2}-1=0,(k<b)E_2$ is inscribed in $E_1$. If $E_1$ and $E_2$ have same eccentricities then length of minor axis of $E_2=p($LLR of $E_1)$ then $p=?$

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{2}$

If eccentricities both are same the

$\begin{array}{c}\frac{b^2}{a^2}+1=\frac{k^2}{b^2}+1\\ ak=b^2\\ k=\frac{b^2}{a}\end{array}$

$\begin{array}{c}\left(Lengthofminoraxisof{E}_2\right)\\ \frac{b^2}{a}=p\left(\frac{2b^2}{a}\right)\\ \left[P=\frac{1}{2}\right]\end{array}$