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

To find the length of the chord of the ellipse
$\frac{x^2}{25}+\frac{y^2}{16}=1$
whose midpoint is $(1,\frac{2}{5})$, we use the formula for the chord with a given midpoint $P(x_1,y_1)$, which is:
$T=S_1$

Here, the equation of the ellipse is:
$S\equiv \frac{x^2}{25}+\frac{y^2}{16}-1=0$
And the midpoint is $(x_1,y_1)=(1,\frac{2}{5})$.

First, we calculate $T$ and $S_1$:
$T=\frac{x\left(1\right)}{25}+\frac{y(2/5)}{16}-1=\frac{x}{25}+\frac{y}{40}-1$
$S_1=\frac{1^2}{25}+\frac{{(2/5)}^2}{16}-1=\frac{1}{25}+\frac{4}{400}-1=\frac{1}{25}+\frac{1}{100}-1=\frac{5}{100}-1=\frac{1}{20}-1$

Equating $T=S_1$:
$\frac{x}{25}+\frac{y}{40}-1=\frac{1}{20}-1\Rightarrow \frac{x}{25}+\frac{y}{40}=\frac{1}{20}$
Multiplying the entire equation by 200 to clear the denominators:
$8x+5y=10\Rightarrow y=\frac{10-8x}{5}$

To find the points of intersection of this line with the ellipse, we substitute $y=\frac{10-8x}{5}$ into the equation of the ellipse:
$\frac{x^2}{25}+\frac{{(10-8x)}^2}{25\times 16}=1$
Multiplying by 400:
$16x^2+{(10-8x)}^2=400$
$16x^2+100-160x+64x^2=400$
$80x^2-160x-300=0$
Dividing by 20:
$4x^2-8x-15=0$

Let the roots of this quadratic equation be $x_a$ and $x_b$, which are the x-coordinates of the endpoints of the chord. Using the quadratic formula:
$x=\frac{8\pm \sqrt{64-4\left(4\right)(-15)}}{8}=\frac{8\pm \sqrt{64+240}}{8}=\frac{8\pm \sqrt{304}}{8}=1\pm \frac{\sqrt{19}}{2}$
The difference between the x-coordinates is:
$|x_b-x_a|=\sqrt{19}$

Since the line is $y=2-\frac{8}{5}x$, the difference between the y-coordinates is:
$|y_b-y_a|=\frac{8}{5}|x_b-x_a|$

The length of the chord $L$ is given by:
$L=\sqrt{{(x_b-x_a)}^2+{(y_b-y_a)}^2}=|x_b-x_a|\sqrt{1+{\left(\frac{8}{5}\right)}^2}$
$L=\sqrt{19}Õ\sqrt{1+\frac{64}{25}}=\sqrt{19}Õ\sqrt{\frac{89}{25}}=\frac{\sqrt{19Õ89}}{5}$
$L=\frac{\sqrt{1691}}{5}$

Thus, the correct option is:
$\frac{\sqrt{1691}}{5}$