QJEEM25Mab3b8: JEE Main 2025 Mathematics - Class 11 JEEM25Mab3b8 Solution: The length of the chord of the ellipse , whose mid-point is ,…

$\begin{aligned}&\begin{aligned}& \mathrm{T}=\mathrm{S}_1 \\& \frac{\mathrm{x} \cdot 1}{4}+\frac{\mathrm{y} \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8} \\& \mathrm{x}+\mathrm{y}=\frac{3}{2}\end{aligned}\\&\text { solve with ellipse }\\&\begin{aligned}\mathrm{PR} & =\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2} \\& =\sqrt{2}\left|\mathrm{x}_2-\mathrm{x}_1\right|\end{aligned}\end{aligned}$

$\begin{aligned}& y_2=\frac{3}{2}-x_2 \\& y_1=\frac{3}{2}-x_1 \\& y_2-y_1=x_2-x_1 \\& x^2+2 y^2=4 \\& x^2+2\left(\frac{3}{2}-x\right)^2=4 \\& 6 x^2-12 x+1=0 \\& x_1+x_2=2 \\& x_1 x_2=1 / 6 \\& \left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2} \\& \quad=\sqrt{4-4 / 6} \\& P R=\sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15}\end{aligned}$