Mathematics · Ellipse

JEE Main 2025 — 4 April, Morning Shift — Question 35

Find the length of latus rectum of an ellipse if foci are $(2,5)$ and $(2,-3)$ and the eccentricity of the ellipse is $\frac{4}{5}$

Answer: C

$F_{1}:(2,5)$ and $F_{2}:(2,-3)$ , notice major axis along $y$ -axis

$\Rightarrow F_{1} F_{2}=8=2 b e \Rightarrow b=\frac{8}{2 e}=\frac{4}{4 / 5}=5$

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

$\Rightarrow a^{2}=9 \Rightarrow a=3$

The length of latus rectum : $\frac{2 a^{2}}{b}=\frac{2(9)}{5}=\frac{18}{5}$

Answer key and solution verified before publishing.

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