# Let x=4 be a directix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If P$(1, \beta) , \beta >0$ is a point on this ellipse, then the equation of the normal to it at P is: Option: 1 Option: 2 Option: 3 Option: 4

$\\\text { Ellipse }: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\ \text { directrix }: x=\frac{a}{e}=4 \;\;\& \;\;e=\frac{1}{2}$

\begin{aligned} &\Rightarrow \mathrm{a}=2 \& \mathrm{~b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)=3\\ &\Rightarrow \text { Ellipse is } \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \end{aligned}

$\mathrm{p} \text { is }\left(1, \frac{3}{2}\right)$

4x - 2y = 1 satisies this point

## Most Viewed Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-