JEE Main 2025 — 28 January, Evening Shift — Question 5

If $\mathrm{midpoint :}\left(\sqrt{2}, \frac{4}{3}\right)$ then equation of AB is $\mathrm{T}=\mathrm{S}_{1}$

$\frac{x \sqrt{2}}{9}+\frac{y}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^{2}}{9}+\frac{\left(\frac{4}{3}\right)^{2}}{4}$

$\frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}$

$\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3}$ 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 $y=2 \quad y=\frac{2}{3}$

Length of chord $=\sqrt{(2 \sqrt{2}-0)^{2}+\left(\frac{2}{3}-2\right)^{2}}$

$$=\sqrt{8+\frac{16}{9}}$$ $$=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22},, \text { so } \alpha=22$$