Question

If length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ whose middle point is $(\frac{1}{2},\frac{2}{5}),$ is $7\sqrt{\frac{k}{25}}$ unit, then $k=$

Answer 1

Equation of the ellipse is :
$\frac{x^2}{25}+\frac{y^2}{16}=1\dots \left(1\right)$
Equation of the chord having $(\frac{1}{2},\frac{2}{5})$ as midpoint is :
$T=S_1$
$\Rightarrow \frac{\left(\frac{1}{4}\right)}{25}+\frac{\left(\frac{4}{25}\right)}{16}-1=\frac{\left(\frac{1}{2}\right)x}{25}+\frac{\left(\frac{2}{5}\right)y}{16}-1$
$\Rightarrow 4x+5y=4$
$\Rightarrow 5y=4(1-x)\dots \left(2\right)$

Solving with equation $\left(1\right),$ we get
$16x^2+16(1-x{)}^2=400$
$\Rightarrow x^2-x-12=0$
$\Rightarrow x=4,-3$
From equation $\left(2\right)$,
For $x=4,y=-\frac{12}{5}$
For $x=-3,y=\frac{16}{5}$
Therefore, length of the chord
$=\sqrt{7^2+{\left(\frac{28}{5}\right)}^2}=7\sqrt{\frac{41}{25}}$ unit
Hence, $k=41$