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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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