Question

A point on the ellipse $4x^2+9y^2=36$, where the normal is parallel to the line $4x-2y-5=0$, is

• A. $\left(\frac{8}{5},\frac{-9}{5}\right)$
• B. $\left(\frac{-9}{5},\frac{8}{5}\right)$
• C. $\left(\frac{9}{5},\frac{8}{5}\right)$
• D. $\left(\frac{8}{5},\frac{9}{5}\right)$

Answer 1

Answer: (C)

Let $P\left(x_1,y_1\right)$ be the point on curve $4x^2+9y^2=36$, where the normal is parallel to line $4x-2y$ SiellisC $=5$. $=5$.
${\therefore \frac{dy}{dx}|}_{P\left(x_1,y_1\right)}=\left(\frac{-4x_1}{9y_1}\right)$
Now, $\frac{9y_1}{4x_1}=2\Rightarrow 9y_1=8x_1$
Also, $4x_1^2+9y_1^2=36$
$\therefore$ On solving (1) and (2), we get
$P\left(x_1=\frac{9}{5},y_1=\frac{8}{5}\right)\text{ or }\left(x_1=\frac{-9}{5},y_1=\frac{-8}{5}\right)$