Question

Find the point on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}$ on which the normals are parellel to the line $2x-y=1$

Answer 1

$\frac{x^2}{4}+\frac{y^2}{9}=1$ $(a=2,b=3)$ major$=b$ as $b>a$.

Let there be a point $P$ on which normal is drawn parallel to $2x-y=1---\left(i\right)$

$\Rightarrow$ slope$=2=m$

Foot of normal: $(\pm \frac{b^2}{\sqrt{b^2+a^2{m}^2}},\pm \frac{m{a}^2}{\sqrt{b^2+a^2{m}^2}})$

$:(\pm \frac{9}{\sqrt{9+4\left(4\right)}},\pm \frac{2\left(4\right)}{\sqrt{9+4\left(4\right)}})$

$:(\pm \frac{9}{5},\pm \frac{8}{5})$