# Get Answers to all your Questions

### Answers (1)

Consider the ellipse $\mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1}$

Equation of normal at any point  $\mathrm{(a \cos \theta, b \sin \theta)}$ is

$\mathrm{a \sin \theta \cdot x-b \cos \theta . y=a^2 e^2 \sin \theta \cos \theta}$             $....(1)$

Take the focus as $\mathrm{S(ae, 0).}$

Since (1) passes through S(a e, 0), we have

$\mathrm{a \sin \theta \cdot a e=a^2 e^2 \sin \theta \cos \theta}$

$\mathrm{\Rightarrow \sin \theta(1-\mathrm{e} \cos \theta)=0}$

$\mathrm{\Rightarrow \sin \theta=0}$ or $\mathrm{\cos \theta=\frac{1}{\mathrm{e}}}$

$\mathrm{\Rightarrow \sin \theta=0 \Rightarrow \theta=0}$ or $\mathrm{\pi \quad(\cos \theta=1 / \mathrm{e}}$ is not possible )

$\Rightarrow$ corresponding points are $\mathrm{(\mathrm{a}, 0)}$ and $\mathrm{(-\mathrm{a}, 0)}$

Similarly, if we take $\mathrm{S^{\prime} \equiv(-a e, 0)}$, then we get same points.

View full answer

## JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE