Question

Let $x=4$be a directrix to an ellipse whose center 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

$4x-2y=1$

Explanation for the correct option:

Finding the equation of normal:

Given;

$e=\frac{1}{2}$

$$\begin{gathered} x=\frac{a}{e}=4 \\ \Rightarrow \frac{a}{e}=4 \\ \Rightarrow \frac{a}{{\displaystyle \frac{1}{2}}}=4 \\ \Rightarrow 2a=4 \\ \Rightarrow a=\frac{4}{2} \\ \Rightarrow a=2 \end{gathered}$$

Finding $b^2$

$$\begin{gathered} e^2=\frac{1-b^2}{a^2} \\ \Rightarrow {\left(\frac{1}{2}\right)}^2=\frac{1-b^2}{2^2}\left[\because e=\frac{1}{2}\right] \\ \Rightarrow \frac{1}{4}=1-\left(\frac{b^2}{4}\right) \\ \Rightarrow \left(\frac{b^2}{4}\right)=\frac{3}{4} \\ \Rightarrow b^2=3 \end{gathered}$$

The general equation of Ellipse is $\left(\frac{x^2}{4}\right)+\left(\frac{y^2}{3}\right)=1$

Given,$P(1,\beta )$

$x=1$

$$\begin{gathered} \left(\frac{1}{4}\right)+\left(\frac{{\beta }^2}{3}\right)=1 \\ \left(\frac{{\beta }^2}{3}\right)=\frac{3}{4} \\ {\beta }^2=\frac{9}{4} \\ \beta =\frac{3}{2} \end{gathered}$$

Therefore, $P\left(1,\frac{3}{2}\right)$

The general equation of normal is $\left(\frac{a^{2}x}{x_1}\right)-\left(\frac{b^{2}y}{y_1}\right)=a^2-b^2$

$$\begin{gathered} \frac{4x}{1}-\frac{3y}{{\displaystyle \frac{3}{2}}}=4-3 \\ \Rightarrow 4x-2y=1 \end{gathered}$$

Therefore, the correct answer is option (B).