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

JEE Main2026Jan 22, Shift 2Mathematics

Let $S$ and $S^{'}$ be the foci of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If $(S P)^{2} + (S^{'} P)^{2} - S P \cdot S^{'} P = 37$ , then $α^{2} + β^{2}$ is equal to :

A
15
B
11
C
17
D
13

Solution

For the ellipse, $a = 5$ , $b = 3$ , $e = \frac{4}{5}$ , so $S P + S^{'} P = 2 a = 10$ .

Given $(S P)^{2} + (S^{'} P)^{2} - S P \cdot S^{'} P = 37$ . Using $(S P + S^{'} P)^{2} = S P^{2} + S^{'} P^{2} + 2 \cdot S P \cdot S^{'} P$ :

$100 - 3 \cdot S P \cdot S^{'} P = 37 \Rightarrow S P \cdot S^{'} P = 21$ .

Focal distances: $S P = a - e α = 5 - \frac{4 α}{5}$ , $S^{'} P = 5 + \frac{4 α}{5}$ .

$S P \cdot S^{'} P = 25 - \frac{16 α ^{2}}{25} = 21 \Rightarrow α^{2} = \frac{25}{4}$ .

From the ellipse equation: $\frac{1}{4} + \frac{β ^{2}}{9} = 1 \Rightarrow β^{2} = \frac{27}{4}$ .

$α^{2} + β^{2} = \frac{25 + 27}{4} = 13$ .

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