Ellipse Question - None

Question

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

Accepted Answer

Let $m$ is midpoint of a chord of the given ellipse

The equation of chord $A B$ with midpoint $m\left(\sqrt{2}, \frac{4}{3} ight)$ is
${T}={S}_{1}$
$\Rightarrow \frac{x \sqrt{2}}{9}+\frac{y}{4}\left(\frac{4}{3} ight)=\frac{(\sqrt{2})^{2}}{9}+\frac{\left(\frac{4}{3} ight)^{2}}{4}$
$\Rightarrow \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}$
$\Rightarrow \frac{\sqrt{2} x+3 y}{9}=\frac{6}{9}$
$\Rightarrow \sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3}$ ...(i)
Substituting into the ellipse equation, we get
$\frac{x^{2}}{9}+\frac{(6-\sqrt{2} x)^{2}}{9 imes 4}=1$
$\Rightarrow 4 x^{2}+36+2 x^{2}-12 \sqrt{2} x=36$
$\Rightarrow 6 x^{2}-12 \sqrt{2} x=0$
$\Rightarrow 6 x(x-2 \sqrt{2})=0$
$\Rightarrow x=0 \& x=2 \sqrt{2}$
Putting these values in equation (i), we get
$y=2, y=\frac{2}{3}$
Hence $A=(0,2)$ and $B=\left(2 \sqrt{2}, \frac{2}{3} ight)$
Length of chord $A B$
$=\sqrt{(2 \sqrt{2}-0)^{2}+\left(\frac{2}{3}-2 ight)^{2}}$
$=\sqrt{8+\frac{16}{9}}$
$=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22}$
Given length of chord $A B=\frac{2 \sqrt{\alpha}}{3}$
Hence, $\alpha=22$

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