Question

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$. If one of its directrices is $\mathrm{x}=-4$ ,then the equation of the normal to it at $\left(1,\frac{3}{2}\right)$ is

• A. $4\mathrm{x}-2\mathrm{y}=1$
• B. $4\mathrm{x}+2\mathrm{y}=7$
• C. $\mathrm{x}+2\mathrm{y}=4$
• D. $2\mathrm{y}-\mathrm{x}=2$

Answer 1

The correct option is A

$4\mathrm{x}-2\mathrm{y}=1$

Finding the equation of normal at $\left(1,\frac{3}{2}\right)$:

Given,

Directrix $\frac{\mathrm{a}}{\mathrm{e}}=-4$ and eccentricity $\mathrm{e}=\frac{1}{2}$

Substituting the value of $e$, we get

$\begin{array}{rcl}\left(\frac{-\mathrm{a}}{\mathrm{e}}\right)& =& -4\\ \left(\frac{-\mathrm{a}}{{\displaystyle \frac{1}{2}}}\right)& =& -4\\ \mathrm{a}& =& 2\end{array}$

Now,

$$\begin{gathered} {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right) \\ =3 \end{gathered}$$

The general equation of ellipse is $\frac{{\mathrm{x}}^2}{a^2}+\frac{{\mathrm{y}}^2}{b^2}=1$.

By substituting the values of $a,b$, we get

$\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$

Equation of normal at $\left(1,\frac{3}{2}\right)$ is

$$\begin{gathered} \frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2 \\ \Rightarrow \frac{4x}{{\displaystyle 1}}-\frac{3y}{{\displaystyle \frac{3}{2}}}=4-3 \\ \Rightarrow 4\mathrm{x}-\frac{6\mathrm{y}}{3}=1 \\ \Rightarrow 4\mathrm{x}-2y=1 \end{gathered}$$

Hence, option (A) is the correct answer.