JEE Main 2025 — Mathematics Coordinate Geometry

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