Ellipse
Two Ellipses Same Eccentricity — Distance Between Foci
nta_pyq_2026_jan
Grade None
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 $\dfrac{4}{5}$. 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
\dfrac{32}{5}
\dfrac{8}{5}
\dfrac{16}{5}
\dfrac{96}{5}
Step-by-Step Solution
Key Concept: Both have $e=\tfrac{4}{5}$. For $E_1$: $2ae=8\Rightarrow a=5$. $b^2=a^2(1-e^2)=9$. $l_1=\tfrac{2b^2}{a}=\tfrac{18}{5}$. For $E_2$: $l_2=\tfrac{2A^2}{B}=\tfrac{18B}{25}$.
$B=4$. Distance $=\tfrac{32}{5}$.
Correct Answer: 1