Discussion Forum

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

Answer:

Explanation:

We have e =$\frac{1}{2}$   and $\frac{a}{e}$=4

$\therefore$    a=2

  Now, $b^{2}=a^{2}(1-e^{2})=(2)^{2}[1-(\frac{1}{2})^{2}]$

= $4(1-\frac{1}{4})=3\Rightarrow b=\sqrt{3}$

$\therefore$   Equation of the ellipse is

            $\frac{x^{2}}{(2)^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1$

$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

  Now, the equation of normal at (1,$\frac{3}{2}$) is

       $\frac{a^{2}x}{x_{1}}-\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$

$\Rightarrow\frac{4x}{1}-\frac{3y}{(3/2)}=4-3$

  $\Rightarrow 4x-2y=1$