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JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Ellipse

The centre of a circle C is at the centre of the ellipse E:...

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 \gt 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

\mathrm{PF}_1 \mathrm{~F}_2

is 30 and the length of the major axis of E is 17 , then the distance between the foci of E is :

Options

A $26$
B $13$
C $12$
D $\frac{13}{2}$

Correct answer

B. $13$

Step-by-step solution

\begin{aligned} & \frac{1}{2} \mathrm{PF}_1 \cdot \mathrm{PF}_2=30 \\ & \mathrm{PF}_1+\mathrm{PF}_2=17 \\ & \mathrm{PF}_1=12 \mathrm{PF}_2=5 \\ & \mathrm{ F}_1 \mathrm{ F}_2=13 \\ & \text{ option }(2)\end{aligned}

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