QJEEM24Maef14: JEE Main 2024 Mathematics - Class 11 JEEM24Maef14 Solution: The length of the chord of the ellipse , whose mid point is ,…

Equation of chord with given middle point.

$\begin{aligned}& T=S_1 \\& \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\& \frac{8 x+5 y}{200}=\frac{8+2}{200} \\& y=\frac{10-8 x}{5} \quad \text{.... (i)}\end{aligned}$

$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$ (put in original equation)

$\begin{aligned}& \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\& 4 x^2-8 x-15=0 \\& x=\frac{8 \pm \sqrt{304}}{8} \\& x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}\end{aligned}$

Similarly, $y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$

$\begin{aligned}& \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\& \text { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\& =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\\end{aligned}$