Question

The equation of normal at point $P(8\sqrt{2},1)$ on the ellipse $\frac{x^2}{144}+\frac{y^2}{9}=1$ is

• A. $\sqrt{2}x+y=15$
• B. $\sqrt{2}x-y=15$
• C. $x+\sqrt{2}y=15$
• D. $x-\sqrt{2}y=15$

Answer 1

The correct option is B $\sqrt{2}x-y=15$

Since point $P(8\sqrt{2},1)$ lies on the ellipse $\frac{x^2}{144}+\frac{y^2}{9}=1$
Using point form of normal :
$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
Here
$a=12,b=3,(x_1,y_1)=(8\sqrt{2},1)\Rightarrow \frac{144x}{8\sqrt{2}}-\frac{9y}{1}=144-9\Rightarrow 9\sqrt{2}x-9y=135\therefore \sqrt{2}x-y=15$