Question
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:
A.
$$x + 2y = 4$$
B.
$$2y - x = 2$$
C.
$$4x - 2y = 1$$
D.
$$4x + 2y = 7$$
Answer :
$$4x - 2y = 1$$
Solution :
Eccentricity of ellipse $$ = \frac{1}{2}$$
$$\eqalign{
& {\text{Now,}}\, - \frac{a}{e} = - 4 \Rightarrow a = 4 \times \frac{1}{2} = 2 \Rightarrow a = 2 \cr
& {\text{We have }}{b^2} = {a^2}\left( {1 - {e^2}} \right) = {a^2}\left( {1 - \frac{1}{4}} \right) = 4 \times \frac{3}{4} = 3 \cr
& \therefore \,{\text{Equation of ellipse is }}\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1 \cr
& {\text{Now differentiating, we get}} \cr
& \Rightarrow \frac{x}{2} + \frac{{2y}}{3} \times y' = 0 \Rightarrow y' = - \frac{{3x}}{{4y}} \cr
& y'\left| {_{\left( {1,\frac{3}{2}} \right)}} \right| = - \frac{3}{4} \times \frac{2}{3} = - \frac{1}{2} \cr
& {\text{Slope}}\,{\text{of}}\,{\text{normal}} = 2 \cr
& \therefore \,\,{\text{Equation}}\,{\text{of}}\,{\text{normal}}\,{\text{at}}\,\left( {1,\frac{3}{2}} \right)\,{\text{is}} \cr
& y - \frac{3}{2} = 2\left( {x - 1} \right) \Rightarrow 2y - 3 = 4x - 4\therefore 4x - 2y = 1 \cr} $$