The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to :

<p>Equation of chord with given middle point.</p> <p>$$\begin{aligned} & T=S_1 \\ & \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\ & \frac{8 x+5 y}{200}=\frac{8+2}{200} \\ & y=\frac{10-8 x}{5} \quad \text{.... (i)} \end{aligned}$$</p> <p>$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$ (put in original equation)</p> <p>$$\begin{aligned} & \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\ & 4 x^2-8 x-15=0 \\ & x=\frac{8 \pm \sqrt{304}}{8} \\ & x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8} \end{aligned}$$</p> <p>Similarly, $y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$</p> <p>$$\begin{aligned} & \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\ & \text { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\ & =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\ \end{aligned}$$</p>