Coordinate Geometry – Ellipse – JEE Main 22 Jan 2026 Shift 2

Solution:

For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$, we have $a^2 = 25 \Rightarrow a = 5$ and $b^2 = 9$.
We know that $SP + S’P = 2a = 10$.

Given equation:
$$ (SP)^2 + (S’P)^2 – SP \cdot S’P = 37 $$
We can rewrite $(SP)^2 + (S’P)^2$ as $(SP + S’P)^2 – 2SP \cdot S’P$.
$$ (SP + S’P)^2 – 2SP \cdot S’P – SP \cdot S’P = 37 $$
$$ (10)^2 – 3 SP \cdot S’P = 37 $$
$$ 100 – 3 SP \cdot S’P = 37 $$
$$ 3 SP \cdot S’P = 63 \Rightarrow SP \cdot S’P = 21 $$

The focal distances are $SP = a – ex_1$ and $S’P = a + ex_1$ (where $x_1 = \alpha$).
$$ SP \cdot S’P = (a – e\alpha)(a + e\alpha) = a^2 – e^2\alpha^2 $$
Calculate eccentricity $e$:
$$ e^2 = 1 – \frac{b^2}{a^2} = 1 – \frac{9}{25} = \frac{16}{25} \Rightarrow e = \frac{4}{5} $$

Substitute values:
$$ 25 – \left(\frac{16}{25}\right)\alpha^2 = 21 $$
$$ \frac{16}{25}\alpha^2 = 4 $$
$$ \alpha^2 = \frac{4 \times 25}{16} = \frac{25}{4} $$

Since P lies on the ellipse:
$$ \frac{\alpha^2}{25} + \frac{\beta^2}{9} = 1 $$
$$ \frac{25/4}{25} + \frac{\beta^2}{9} = 1 $$
$$ \frac{1}{4} + \frac{\beta^2}{9} = 1 $$
$$ \frac{\beta^2}{9} = \frac{3}{4} \Rightarrow \beta^2 = \frac{27}{4} $$

We need $\alpha^2 + \beta^2$:
$$ \alpha^2 + \beta^2 = \frac{25}{4} + \frac{27}{4} = \frac{52}{4} = 13 $$

Ans. (4)