Question

Sum of distance's from the $x-$axis to the point(s) on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1,$ where the normal is parallel to the line $2x+y=1,$ is $\frac{k}{5}$ unit, then $k=$

Answer 1

Equation of normal at $(3\cos \theta ,2\sin \theta )$ is
$3x\sec \theta -2y\text{cosec}\theta =5$
$y=\frac{3}{2}\left(\tan \theta \right)x-\frac{5\sin \theta }{2}$

If the normal is parallel to the given line $2x+y=1,$ then
$\frac{3}{2}\left(\tan \theta \right)=-2$
$\Rightarrow \cos \theta =\mp \frac{3}{5},\sin \theta =\pm \frac{4}{5}$

Hence, points are $(\frac{-9}{5},\frac{8}{5})$ and $(\frac{9}{5},-\frac{8}{5})$

By distance formula, the required sum of distance
$=\frac{16}{5}$ unit