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

If $\mathrm{m}\left(\sqrt{2},\frac{4}{3}\right)$  than equation of of AB is

$\begin{array}{l}\mathrm{T}={\mathrm{S}}_1\\ \frac{\mathrm{x}\sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2}{)}^2}{9}+\frac{{\left(\frac{4}{3}\right)}^2}{4}\\ \frac{\sqrt{2}\mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9}\end{array}$

$\sqrt{2}x+3y=6\Rightarrow y=\frac{6-\sqrt{2}x}{3}$ put in ellipse
So, $\frac{x^2}{9}+\frac{(6-\sqrt{2}x{)}^2}{9\times 4}=1$

$\begin{array}{l}4x^2+36+2x^2-12\sqrt{2}x=36\\ 6x^2-12\sqrt{2}x=0\\ 6x(x-2\sqrt{2})=0\\ x=0\mathrm{\&}x=2\sqrt{2}\\ \text{ So }y=2y=\frac{2}{3}\end{array}$

$\begin{array}{r}\text{ Length of chord }=\sqrt{(2\sqrt{2}-0{)}^2+{\left(\frac{2}{3}-2\right)}^2}\\ =\sqrt{8+\frac{16}{9}}\\ =\sqrt{\frac{88}{9}}=\frac{2}{3}\sqrt{22}\text{ so }\alpha =22\end{array}$