Question

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0,(a>b)$ and $E_2:\frac{x^2}{k^2}+\frac{y^2}{b^2}-1=0,(k<b)$ is inscribed in $E_1.$ If $E_1\text{and}{E}_2$ have same eccentricities and length of minor axis of $E_2=p\times \text{Length of Latus rectum of}{E}_1,\text{then}p=$

• A. 1.0
• B. 1.00
• C. 1

Answer 1

Answer: (A), (B), (C)

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0,(a>b)$ and $E_2:\frac{x^2}{k^2}+\frac{y^2}{b^2}-1=0,(k<b)$

Let $e_1$ and $e_2$ are the eccentricities of ellipse $E_1$ and $E_2,$ then given that,
$e_1=e_2\Rightarrow e_1^2=e_2^2\Rightarrow 1-\frac{b^2}{a^2}=1-\frac{k^2}{b^2}$
$\Rightarrow k=\frac{b^2}{a}\Rightarrow 2k=\frac{2b^2}{a}\Rightarrow \text{length of minor axis of}{E}_2=1\times \left[\text{LLR of}{E}_1\right]\therefore p=1$