Let each of the two ellipses E_1:x^2a^2+y^2b^2=1,(ab) and E_2:x^2A^2+y^2B^2=1,(A

Question

Let each of the two ellipses ${E}_{1}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,(a>b)$ and ${E}_{2}:\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1,(A<B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of ${E}_{1}$ and ${E}_{2}$ be ${l}_{1}$ and ${l}_{2}$, respectively, such that $2{l}_{2}^{1}=9{l}_{2}$. If the distance between the foci of ${E}_{1}$ is 8 , then the distance between the foci of ${E}_{2}$ is

Accepted Answer

Given, for ellipse ${E}_{1}$ and ${E}_{2}$ eccentricity is $\frac{4}{5}$.

${E}_{1}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\, a>b$

distance between foci $=8$

$2ae=8\implies ae=4$

$\Rightarrow$ $a\times \frac{4}{5}=4\implies a=5$

$\Rightarrow$ $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}=\frac{4}{5}$

$\Rightarrow$ $1-\frac{{b}^{2}}{{5}^{2}}=\frac{16}{25}$

$\Rightarrow$ $\frac{{b}^{2}}{{5}^{2}}=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow$ $b=3$

Length of latus rectum for ${E}_{1},{l}_{1}=\frac{2{b}^{2}}{a}$

${l}_{1}=2\times \frac{{3}^{2}}{5}=\frac{18}{5}$

For ${E}_{2}:\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1\, (A$ $$\& e=\frac{4}{5}  \  \& A$1-\frac{{A}^{2}}{{B}^{2}}=\frac{16}{25}$$$

$\Rightarrow$ $\frac{{A}^{2}}{{B}^{2}}=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow$ ${A}^{2}=\frac{9}{25}{B}^{2}$

If ${l}_{2}$ is latus rectum of ${E}_{2},{l}_{2}=\frac{2{A}^{2}}{B}$

${l}_{2}=\frac{2}{B}(\frac{9}{25}{B}^{2})=\frac{18}{25}B$

It is given that $2{l}_{2}^{1}=9{l}_{2}$

$2{(\frac{18}{5})}^{2}=9\times \frac{18}{25}B$

$\Rightarrow$ $2\times \frac{18}{9}=B$

$\Rightarrow$ $B=4$

distance between foci of ${E}_{2}$ is $2Be$

$=2\times 4\times \frac{4}{5}=\frac{32}{5}$

Solutions · 1