If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$, then $\alpha$ is :

Question

Aicharya · Accepted Answer

$\begin{aligned}
& E: \frac{x^2}{9}+\frac{y^2}{4}=1 \\ & T=S_1 \\ & \Rightarrow \frac{\sqrt{2} x}{9}+\frac{1}{4}\left(\frac{4}{3} y\right)-1=\frac{2}{9}+\frac{16}{9(4)}-1 \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9} \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{3} \quad \Rightarrow \sqrt{2 x}+3 y=6
\end{aligned}$
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 $\alpha=22$

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the…

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