Mathematics - Ellipse Question with Solution | TestHub

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