Question
Let each of the two ellipses $E_{1}:\frac{x^{2}}{a^{2}} + \frac{y^{2}}{{\text{ }b}^{2}} = 1,(a > b)$ and $E_{2}:\frac{x^{2}}{{\text{ }A}^{2}} + \frac{y^{2}}{{\text{ }B}^{2}} = 1,(\text{ }A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_{1}$ and $E_{2}$ be $\mathcal{l}_{1}$ and $\mathcal{l}_{2}$, respectively, such that $2\mathcal{l}_{1}^{2} = 9\mathcal{l}_{2}$. If the distance between the foci of $E_{1}$ is 8 , then the distance between the foci of $E_{2}$ is
Options
Solution
$2ae = 8 \Rightarrow a = 5$
$${b^{2} = a^{2}\left( 1 - e^{2} \right) }{b^{2} = a^{2} \times \frac{9}{25}\ {\text{ }b}^{2} = 9 }{E_{1}:\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 }{\mathcal{l}_{1}:\frac{2{\text{ }b}^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5} }{A^{2} = B^{2}\left( 1 - e^{2} \right) \Rightarrow A^{2} = \frac{9}{25}B^{2} \Rightarrow A = \frac{3}{5}B }{2\mathcal{l}^{2} = 9\mathcal{l}_{2} \Rightarrow 2\left( \frac{18}{5} \right)^{2} = 9\mathcal{l}_{2} \Rightarrow \mathcal{l}_{2} = \frac{4 \times 18}{25} }{\frac{2{\text{ }A}^{2}}{\text{ }B} = \frac{72}{25} \Rightarrow {\text{ }A}^{2}\frac{36}{25}\text{ }B }$$$\frac{9}{25}{\text{ }B}^{2} = \frac{36\text{ }B}{25} \Rightarrow \text{ }B = 4$,
Distance between focii $2Be = 2 \times \frac{4}{5} \times 4 = \frac{32}{5}$
Create a free account to view solution
View Solution Free