JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $P (10, 215)$ be a point on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , whose foci are $S$ and $S^{'}$ . If the length of its latus rectum is 8, then the square of the area of $Δ P S S^{'}$ is equal to :

A
4200
B
900
C
1462
D
2700

Solution

$P$ lies on the hyperbola: $\frac{100}{a ^{2}} - \frac{60}{b ^{2}} = 1$ … (1).

Latus rectum $= \frac{2 b ^{2}}{a} = 8 \Rightarrow b^{2} = 4 a$ … (2).

Substitute into (1): $\frac{100}{a ^{2}} - \frac{60}{4 a} = 1 \Rightarrow 400 - 15 a \cdot a = a^{2}$ ... Cleaning up: $4 a^{2} + 60 a - 400 = 0 \Rightarrow a^{2} + 15 a - 100 = 0 \Rightarrow a = 5$ (rejecting negative).

So $a = 5$ , $b^{2} = 20$ . Then $e^{2} = 1 + \frac{b ^{2}}{a ^{2}} = 1 + \frac{4}{5} = \frac{9}{5}$ .

Focal length $∣ S S^{'} ∣ = 2 a e = 2 a^{2} e^{2} = 2 25 \cdot 9/5 = 245 = 65$ .

Area of $Δ P S S^{'} = \frac{1}{2} ∣ S S^{'} ∣ \cdot ∣ y_{P} ∣ = \frac{1}{2} \cdot 65 \cdot 215 = 675 = 303$ .

Square of area $= 900 \cdot 3 = 2700$ .

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