JEE Main 2025 Apr 4 Shift 2, Mathematics Q17: Introduction to Ellipse

JEE Main2025Apr 4, Shift 2Mathematics

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 :

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

Solution

Since circle $C$ passes through both foci and is centred at the centre of the ellipse, $F_{1} F_{2}$ is a diameter of $C$ . By Thales' theorem, the angle $∠ F_{1} P F_{2} = 9 0^{\circ}$ at any point $P$ on $C$ .

So triangle $P F_{1} F_{2}$ is right-angled at $P$ . Using the focal-radii property $P F_{1} + P F_{2} = 2 a = 17$ , and the right angle:

area $= \frac{1}{2} P F_{1} \cdot P F_{2} = 30 \Rightarrow P F_{1} \cdot P F_{2} = 60$ .

So $P F_{1}$ and $P F_{2}$ are roots of $u^{2} - 17 u + 60 = 0$ , giving ${5, 12}$ .

By Pythagoras, $F_{1} F_{2} = 5^{2} + 1 2^{2} = 169 = 13$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q17: Introduction to Ellipse

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