Mathematics · Ellipse

JEE Main 2025 — 4 April, Evening Shift — Question 28

The centre of a circle $C$ is at the centre of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ . Let $C$ pass through the

foci $F_{1}$ and $F_{2}$ of $E$ such that the circle $C$ and the ellipse $E$ intersect at four points. Let $P$ be one of these four

points. If the area of the triangle $P F_{1} F_{2}$ is 30 and the length of the major axis of $E$ is 17 , then the distance

between the foci of $E$ is:

Answer: A

$x^{2}+\frac{a^{2} y^{2}}{b^{2}}=a^{2}$

$\Rightarrow y^{2}\left(1-\frac{a^{2}}{b^{2}}\right)=a^{2}\left(e^{2}-1\right)=a^{2}\left(1-\frac{b^{2}}{a^{2}}-1\right)$

$=-b^{2}$

$\Rightarrow \frac{y^{2}\left(b^{2}-a^{2}\right)}{b^{2}}=-b^{2} \Rightarrow y^{2}=\frac{b^{4}}{\left(a^{2}-b^{2}\right)}$

Height $=|y|=\frac{b^{2}}{\sqrt{a^{2}-b^{2}}}$

Area $=(2 a e) \times \frac{1}{2} \times \frac{b^{2}}{\sqrt{a^{2}-b^{2}}}=30$

$=\frac{a b^{2} e}{a \sqrt{1-\frac{b^{2}}{a^{2}}}}=b^{2}, a=\frac{17}{2}$

Distance between foci $=2 a e$

$=17 \sqrt{1-\frac{b^{2}}{a^{2}}}=17 \sqrt{1-\frac{30 \times 4}{289}}=13$

Answer key and solution verified before publishing.

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