Question

Let $x=4$ be a directrix 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

• A. $8x-2y=5$
• B. $4x-2y=1$
• C. $7x-4y=1$
• D. $4x-3y=2$

Answer 1

The correct option is B $4x-2y=1$

Given: $e=\frac{1}{2}$
We know that directrix to an ellipse whose centre is at the origin is
$x=\frac{a}{e}=4$
$\Rightarrow a=2$
and $e^2=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{1}{4}=1-\frac{b^2}{4}$
$\Rightarrow \frac{b^2}{4}=\frac{3}{4}$
$\Rightarrow b^2=3$
Ellipse : $\frac{x^2}{4}+\frac{y^2}{3}=1$
Since $P(1,\beta )$ is a point on this ellipse.
$\therefore \frac{1}{4}+\frac{{\beta }^2}{3}=1$
$\Rightarrow \beta =\frac{3}{2}$
$\therefore P(1,\frac{3}{2})$
Now, equation of normal at point $P(1,\frac{3}{2})$
$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
$\Rightarrow \frac{4x}{1}-\frac{3y}{3/2}=4-3$
$\therefore 4x-2y=1$