Aspire Faculty ID #18688 · Topic: NIMCET 2026 · Just now
NIMCET 2026

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

Solution

For ellipse, 
$e=\frac{1}{2}$ 
Directrix is $x=-\frac{a}{e}$ 

Given, $-\frac{a}{e}=-4$ 
$\frac{a}{e}=4$ 
$a=4e=4\cdot \frac{1}{2}=2$ 
Now, $b^2=a^2(1-e^2)$ 
$b^2=4\left(1-\frac{1}{4}\right)$ 
$b^2=3$ 
So ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$ 

Equation of normal to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ is $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$ 

Here, $a^2=4,\ b^2=3,\ x_1=1,\ y_1=\frac{3}{2}$ 

So, $\frac{4x}{1}-\frac{3y}{3/2}=4-3$ 
$4x-2y=1$ 
Answer: $4x-2y=1$
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