Question

If the normal at P on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ cuts the major and minor axes in Q and R respectively the PQ:PR =

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Equation of the normal at $P(x_1,y_1)$ is $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

Normal cuts major axis at $Q(\frac{a^2-b^2}{a^2}{x}_1,0)$ and minor axis at $(0,\frac{-a^2-b^2}{b^2}{y}_1)$
Now $P{Q}^2=$ ${(\frac{a^2-b^2}{a^2}{x}_1-x_1)}^2+y_1^2=\frac{b^4}{a^4}{x}_1^2+y_1^2=\frac{b^4{x}_1^2+a^4{y}_1^2}{b^4}$
$\frac{P{Q}^2}{P{R}^2}=\frac{b^4}{a^4}\Rightarrow \frac{PQ}{PR}=\frac{b^2}{a^2}$