JEE Main 2025 — Ellipse Question with Solution

$\begin{aligned} & E: \frac{x^2}{9}+\frac{y^2}{4}=1 \\ & T=S_1 \\ & \Rightarrow \frac{\sqrt{2} x}{9}+\frac{1}{4}\left(\frac{4}{3} y\right)-1=\frac{2}{9}+\frac{16}{9(4)}-1 \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9} \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{3} \quad \Rightarrow \sqrt{2 x}+3 y=6 \end{aligned}$ Now point of intersection of chord and ellipse is $\begin{aligned} & \frac{(6-3 y)^2}{18}+\frac{y^2}{4}=1 \\ & \frac{(2-y)^2}{2}+\frac{y^2}{4}=1 \\ & 2\left(4+y^2-4 y\right)+y^2=4 \\ & \Rightarrow 3 y^2-8 y+4=0 \\ & \Rightarrow y=2, \frac{2}{3} \end{aligned}$ So, points are $(0,2)$ are $\left(2\sqrt{2},\frac{2}{3}\right)$ Length of chord $=\sqrt{(2\sqrt{2}{)}^2+{\left(\frac{2}{3}-2\right)}^2}$ $\begin{aligned} & =\sqrt{8+\frac{16}{9}} \\ & =\frac{\sqrt{88}}{3}=\frac{2 \sqrt{22}}{3} \end{aligned}$ On comparing $\alpha =22$