Length of the chord of an ellipse whose midpoint is given Find the length of the chord of the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1\tag1$$where mid point is $(\frac{1}{2},\frac{2}{3})$
My attempt: I know the equation of the chord of an ellipse whose mid-point is given is $T=S_1$,where T is $\left(\frac{x.x_1}{25}+\frac{y.y_1}{16} - 1\right)$ and $S_1$ is $\left(\frac{x_1^2}{25}+\frac{y_1^2}{16} - 1\right)$. Using this I get the equation of the chord as $$\frac{x}{50}+\frac{y}{24}=\frac{17}{450}\tag2$$ Now by solving (1) & (2), I should get the point of intersections and eventually the length of the chord. I want to know if there is a better way to solve this question.

Line passing through midpoint $(\tfrac{1}{2}, \tfrac{2}{3})$ is
$$p(r)=\langle\tfrac{1}{2} + r \cos t, \tfrac{2}{3} + r \sin t\rangle \tag 1$$
 where $r$ is the distance along the line and $t$ is the angle of slope of line.
Using your equation of chord, we can calculate $\cos t$ and $\sin t$:
$$\frac{x}{50}+\frac{y}{24}=\frac{17}{450} \tag 2$$
$\tan t =\frac{-24}{50} = \frac{-12}{25}$, so $\sin t= \frac{12}{\sqrt{25^2+12^2}}$ and $\cos t  = \frac{-25}{\sqrt{25^2 + 12^2}}$
Substitute $\cos t$ and $\sin t$ in equation $1$, substitute $p(r)$ in the ellipse equation and solve for $r$. This gives a quadratic in $r$:
$$r^2 = \dfrac{332977}{15300}$$
So the length of chord is $2r$, ie
$$2r = \frac{1}{15}\sqrt{\dfrac{332977}{17}} \approx 9.3302$$
