JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 4)
Explanation
$$\begin{aligned} &\text { If } \mathrm{m}\left(\sqrt{2}, \frac{4}{3}\right) \text { than equation of of } \mathrm{AB} \text { is }\\ &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4} \\ & \frac{\sqrt{2} \mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3} \text { put in ellipse }\\ &\begin{aligned} & \text { So, } \frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1 \\ & 4 x^2+36+2 x^2-12 \sqrt{2} x=36 \\ & 6 x^2-12 \sqrt{2} x=0 \\ & 6 x(x-2 \sqrt{2})=0 \\ & x=0 \& x=2 \sqrt{2} \\ & \text { So } y=2 \quad y=\frac{2}{3} \\ & \text { Length of chord }=\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2} \\ & \quad=\sqrt{8+\frac{16}{9}} \\ & \quad=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text { so } \alpha=22 \end{aligned} \end{aligned}$$
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