Question

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

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

Answer 1

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

Eccentricity, $e=\frac{1}{2}$
Let $2a\text{and}2b$ be the length of the major-axis and minor-axis respectively.
$\Rightarrow \frac{a}{e}=4\Rightarrow a=2$
$e=\frac{1}{2}\Rightarrow \sqrt{1-\frac{b^2}{a^2}}=\frac{1}{2}$
$\Rightarrow b=\sqrt{3}$
$\therefore$ Equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
Slope of tangent at $(1,\frac{3}{2})\text{is}\frac{dy}{dx}{|}_{(1,3/2)}=-\frac{3\times 1}{4\times \frac{3}{2}}=-\frac{1}{2}$
Thus, equation of normal at $(1,\frac{3}{2})$ is
$y-\frac{3}{2}=2(x-1)$
$\Rightarrow 4x-2y=1$