JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Ellipse

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}

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

PF_{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: (B)

Solution

Because the circle is centred at the centre of the ellipse and passes through both foci,

F_{1} F_{2}

is a diameter, so

∠ F_{1} P F_{2} = 9 0^{\circ}

. Then

P F_{1} \cdot P F_{2} = 2 \times 30 = 60

, while

P F_{1} + P F_{2} = 2 a = 17

. Hence

F_{1} F_{2}^{2} = (P F_{1} + P F_{2})^{2} - 2 P F_{1} \cdot P F_{2} = 289 - 120 = 169

, giving the distance between the foci

F_{1} F_{2} = 13

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option