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 $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1\ (A<B)$ have eccentricity $\dfrac45$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $l_1$ and $l_2$ respectively, such that $2l_1^2=9l_2$. If the distance between the foci of $E_1$ is $8$, then the distance between the foci of $E_2$ is:

Accepted Answer

For $E_1$: $2ae=8\Rightarrow ae=4$, and $e=\dfrac45\Rightarrow a=5$. Then $b^2=a^2(1-e^2)=25\times\dfrac{9}{25}=9\Rightarrow b=3$.

$$l_1=\frac{2b^2}{a}=\frac{2\times9}{5}=\frac{18}{5}.$$

For $E_2$ ($B$ is the semi-major axis since $A<B$): $e=\sqrt{1-\dfrac{A^2}{B^2}}=\dfrac45 \Rightarrow \dfrac{A^2}{B^2}=\dfrac{9}{25}$.

$$l_2=\frac{2A^2}{B}=\frac{2}{B}\cdot\frac{9}{25}B^2=\frac{18B}{25}.$$

Using $2l_1^2=9l_2$:

$$2\left(\frac{18}{5}\right)^2=9\times\frac{18B}{25} \;\Rightarrow\; \frac{648}{25}=\frac{162B}{25} \;\Rightarrow\; B=4.$$

Distance between foci of $E_2 = 2Be = 2\times4\times\dfrac45=\dfrac{32}{5}$.

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