QJEEM25Ma6749: JEE Main 2025 Mathematics - Class 11 JEEM25Ma6749 Solution: If the midpoint of a chord of the ellipse is , and the length…

$\begin{aligned}&\text { If } \mathrm{m}\left(\sqrt{2}, \frac{4}{3}\right) \text { than equation of of } \mathrm{AB} \text { is }\\&\begin{aligned}& \mathrm{T}=\mathrm{S}_1 \\& \frac{\mathrm{x} \sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4} \\& \frac{\sqrt{2} \mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9}\end{aligned}\end{aligned}$

$\begin{aligned}&\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3} \text { put in ellipse }\\&\begin{aligned}& \text { So, } \frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1 \\& 4 x^2+36+2 x^2-12 \sqrt{2} x=36 \\& 6 x^2-12 \sqrt{2} x=0 \\& 6 x(x-2 \sqrt{2})=0 \\& x=0 \& x=2 \sqrt{2} \\& \text { So } y=2 \quad y=\frac{2}{3} \\& \text { Length of chord }=\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2} \\& \quad=\sqrt{8+\frac{16}{9}} \\& \quad=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text { so } \alpha=22\end{aligned}\end{aligned}$