10-2. Parabola, Ellipse, Hyperbola
hard

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 :

A$18$
B$22$
C$26$
D$20$
(JEE MAIN-2025)

Solution

image
If $m \left(\sqrt{2}, \frac{4}{3}\right)$ than equation of of $AB$ is
$T=S_1$
$\frac{x \sqrt{2}}{9}+\frac{y}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4}$
$\frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}$
$\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3}$ put in ellipse
So, $\frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1$
$4 x^2+36+2 x^2-12 \sqrt{2} x=36$
$6 x^2-12 \sqrt{2} x=0$
$6 x(x-2 \sqrt{2})=0$
$x=0 \ x=2 \sqrt{2}$
So $y =2 y =\frac{2}{3}$
$\text { Length of chord } =\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2}$
$ =\sqrt{8+\frac{16}{9}}$
$ =\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text { so } \alpha=22$
Standard 11
Mathematics

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