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 :

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Answer: A. \frac{\sqrt{1691}}{5} 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 	ext{.... (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} \ & 	ext { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2ight)^2+\left(\mathrm{y}_1-\mathrm{y}_2ight)^2} \ & =\sqrt{\frac{4 	imes 304}{64}+\frac{4 	imes 304}{25}}=\frac{\sqrt{1691}}{5} \ \end{aligned}

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 :

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