JEE Main 2025 Jan 24 Shift 2, Mathematics Q1: Ellipse: Chord, Tangent and Normal

For a conic, the chord with a given midpoint $(x_1,y_1)$ satisfies $T=S_1$, where $T$ is the polar form and $S_1$ is the conic evaluated at $(x_1,y_1)$.

For the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ with midpoint $(3,1)$:

$T:\frac{x\cdot 3}{25}+\frac{y\cdot 1}{16}-1=\frac{9}{25}+\frac{1}{16}-1=S_1$

The constant cancels from both sides, giving:

$\frac{3x}{25}+\frac{y}{16}=\frac{9}{25}+\frac{1}{16}$

Multiply through by $400$ (= LCM of 25 and 16):

$48x+25y=144+25=169$

Correct option: (1)