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

Equation of chord with given middle point is given by,

$\mathrm{T}={\mathrm{S}}_1$

$\Rightarrow \frac{\mathrm{x}}{25}+\frac{\mathrm{y}}{40}=\frac{1}{25}+\frac{1}{100}$

$\Rightarrow \frac{8\mathrm{x}+5\mathrm{y}}{200}=\frac{8+2}{200}$

$\Rightarrow \mathrm{y}=\frac{10-8\mathrm{x}}{5} ..\left(\mathrm{i}\right)$

$\Rightarrow \frac{{\mathrm{x}}^2}{25}+\frac{(10-8\mathrm{x}{)}^2}{400}=1$

$\Rightarrow \frac{16{\mathrm{x}}^2+100+64{\mathrm{x}}^2-160\mathrm{x}}{400}=1$

$\Rightarrow 4{\mathrm{x}}^2-8\mathrm{x}-15=0$

$\Rightarrow \mathrm{x}=\frac{8\pm \sqrt{304}}{8}$

$\Rightarrow {\mathrm{x}}_1=\frac{8+\sqrt{304}}{8}; {\mathrm{x}}_2=\frac{8-\sqrt{304}}{8}$

Similarly, $\mathrm{y}=\frac{10-8\pm \sqrt{304}}{5}=\frac{2\pm \sqrt{304}}{5}$

$\Rightarrow {\mathrm{y}}_1=\frac{2-\sqrt{304}}{5};{\mathrm{y}}_2=\frac{2+\sqrt{304}}{5}$

Distance $=\sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}$

$=\sqrt{\frac{4\times 304}{64}+\frac{4\times 304}{25}}$

$=\frac{\sqrt{1691}}{5}$