Question

If the midterm of the chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ is $(0,3)$ and the length of the chord is $\frac{4k}{5},$ then $k$ is,...................

Answer 1

Ellipse: $\frac{x^2}{16}+\frac{y^2}{25}=1---\left(i\right)$

midpoint of chord:$(0,3)$

now equation of chord whose midpoint is given$\Rightarrow S_1=S_{11}$

$\Rightarrow \frac{0\left(x\right)}{16}+\frac{3y}{25}-1=\frac{9}{25}-1$

$y=3---\left(ii\right)$

now point of intersection of $i$ and $ii$,

At $y=3$, $\frac{x^2}{16}+\frac{9}{25}=1$

$\Rightarrow x^2=\frac{(16{)}^2}{25}$

$\Rightarrow x=\pm \frac{16}{5}$

now, distance between two points $(\frac{16}{5},3)$ and $(\frac{-16}{5},3)$

$\frac{4k}{5}=\sqrt{(3-3{)}^2+[\frac{16}{5}-(-\frac{16}{5}){]}^2}$

$\frac{4k}{5}=\sqrt{(\frac{2\left(16\right)}{5}{)}^2}$

$\frac{4k}{5}=\frac{32}{5}$

$\Rightarrow k=8$