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The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directices is $x = - 4$ then the equation of the normal to it at $\left( {1,\frac{3}{2}} \right)$ is
$x + 2y = 4$
$2y - x = 2$
$4x - 2y = 1$
$4x + 2y = 7$
Solution
Eccentricity of ellipse $ = \frac{1}{2}$
Now, $ – \frac{a}{e} = – 4 \Rightarrow a = 4 \times \frac{1}{2} = 2 \Rightarrow a = 2$
we have ${b^2} = {a^2}\left( {1 – {e^2}} \right) = {a^2}\left( {1 – \frac{1}{4}} \right) = 4 \times \frac{3}{4} = 3$
$\therefore $ Equation of ellipse
$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1$
Now differentiating, we get
$ \Rightarrow \frac{x}{2} + \frac{{2y}}{3} \times y' = 0 \Rightarrow y' = – \frac{{3x}}{{4y}}$
$y'\left| {_{\left( {1,3/2} \right)}} \right| = – \frac{3}{4} \times \frac{2}{3} = – \frac{1}{2}$
Slope of normal $=2$
$\therefore $ Equation of normal at $\left( {1,\frac{3}{2}} \right)$ is
$y – \frac{3}{2} = 2\left( {x – 1} \right) \Rightarrow 2y – 4x – 4$
$\therefore 4x – 2y = 1$
