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Chord length of ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1_{\text {with mid point }}$ $\left(1, \frac{2}{5}\right)_{\text {is equal to }}$ $\qquad$
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The correct answer is:
$\frac{\sqrt{1691}}{5}$
Correct option is (2) $\frac{\sqrt{1691}}{5}$

$\begin{aligned} & T=S_1 \Rightarrow \frac{x}{25}+\frac{y \cdot \frac{2}{5}}{16}=\frac{1}{25}+\frac{4}{16 \times 25} \\ & \frac{x}{25}+\frac{y}{8 \times 5}=\frac{5}{100} \\ & 4 x+\frac{5}{2} y=5 \Rightarrow 8 x+5 y-10=0 \\ & \frac{x^2}{25}+\frac{y^2}{16}=1 \Rightarrow(4 x)^2+(5 y)^2=400 \\ & \Rightarrow(4 x)^2+(8 x-10)^2=400 \\ & \Rightarrow 4 x^2+(4 x-5)^2=100 \quad \Rightarrow 20 x^2-40 x-75=0 \\ & \Rightarrow 4 x^2-8 x-15=0\end{aligned}$
$\begin{aligned} & L-\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}-\sqrt{\left(x_2-x_1\right)^2+\left\{\frac{10-8 x_2}{5}-\frac{\left(10-8 x_1\right)}{5}\right\}^2} \\ & =\sqrt{\left(x_2-x_1\right)^2+\frac{64}{25}\left(x_2-x_1\right)^2}=\sqrt{\frac{89}{4}} \sqrt{\left(x_2-x_1\right)^2} \\ & =\sqrt{\frac{89}{4}+\sqrt{\left(x_1-x_2\right)^2-4 x_1 x_2}}=\sqrt{\frac{89}{5}} \sqrt{4+4 \times \frac{15}{4}}=\frac{\sqrt{89} \times \sqrt{19}}{5} \\ & =\frac{\sqrt{1691}}{5}\end{aligned}$

$\begin{aligned} & T=S_1 \Rightarrow \frac{x}{25}+\frac{y \cdot \frac{2}{5}}{16}=\frac{1}{25}+\frac{4}{16 \times 25} \\ & \frac{x}{25}+\frac{y}{8 \times 5}=\frac{5}{100} \\ & 4 x+\frac{5}{2} y=5 \Rightarrow 8 x+5 y-10=0 \\ & \frac{x^2}{25}+\frac{y^2}{16}=1 \Rightarrow(4 x)^2+(5 y)^2=400 \\ & \Rightarrow(4 x)^2+(8 x-10)^2=400 \\ & \Rightarrow 4 x^2+(4 x-5)^2=100 \quad \Rightarrow 20 x^2-40 x-75=0 \\ & \Rightarrow 4 x^2-8 x-15=0\end{aligned}$
$\begin{aligned} & L-\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}-\sqrt{\left(x_2-x_1\right)^2+\left\{\frac{10-8 x_2}{5}-\frac{\left(10-8 x_1\right)}{5}\right\}^2} \\ & =\sqrt{\left(x_2-x_1\right)^2+\frac{64}{25}\left(x_2-x_1\right)^2}=\sqrt{\frac{89}{4}} \sqrt{\left(x_2-x_1\right)^2} \\ & =\sqrt{\frac{89}{4}+\sqrt{\left(x_1-x_2\right)^2-4 x_1 x_2}}=\sqrt{\frac{89}{5}} \sqrt{4+4 \times \frac{15}{4}}=\frac{\sqrt{89} \times \sqrt{19}}{5} \\ & =\frac{\sqrt{1691}}{5}\end{aligned}$
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