(A) $\dfrac{96}{5}$

(B) $\dfrac{8}{5}$

(C) $\dfrac{16}{5}$

(D) $\dfrac{32}{5}$

Correct Answer: $\dfrac{32}{5}$

Explanation

For an ellipse, eccentricity $e=\dfrac{c}{a}$, where $2c$ is the distance between the foci.

Given for both ellipses:

$$ e=\frac45 $$

Ellipse $E_1$

Distance between foci is $8$, hence

$$ 2c_1=8 \Rightarrow c_1=4 $$

Using $e=\dfrac{c_1}{a}$:

$$ \frac45=\frac{4}{a} \Rightarrow a=5 $$

Now,

$$ b^2=a^2(1-e^2)=25\left(1-\frac{16}{25}\right)=9 $$

Length of latus rectum:

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

Ellipse $E_2$

Given relation:

$$ 2l_1^2=9l_2 $$

Substitute $l_1=\dfrac{18}{5}$:

$$ 2\left(\frac{18}{5}\right)^2=9l_2 $$

$$ l_2=\frac{72}{25} $$

For $E_2$, since major axis is along $y$-direction:

$$ l_2=\frac{2A^2}{B} $$

Also,

$$ e=\frac{c_2}{B}=\frac45 \Rightarrow c_2=\frac45 B $$

Using $A^2=B^2(1-e^2)=\frac{9}{25}B^2$:

$$ l_2=\frac{2\times \frac{9}{25}B^2}{B}=\frac{18}{25}B $$

Equate with $\dfrac{72}{25}$:

$$ \frac{18}{25}B=\frac{72}{25} $$

$$ B=4 $$

Thus,

$$ c_2=\frac45\times4=\frac{16}{5} $$

Distance between the foci of $E_2$:

$$ 2c_2=\boxed{\frac{32}{5}} $$

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