Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is ${1 \over 2}$. If P(1, $\beta$), $\beta$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :

$e = {1 \over 2}$ <br><br>$x = {a \over e} = 4$<br><br>$\Rightarrow$ a = 2<br><br>${e^2} = 1 - {{{b^2}} \over {{a^2}}}$ <br><br>$\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}$<br><br>${{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3$<br><br>$\therefore$ Ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$<br><br>P(1, $\beta$) is on the ellipse<br><br> ${1 \over 4} + {{{\beta ^2}} \over 3} = 1$<br><br>${{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta = {3 \over 2}$<br><br>$\Rightarrow P\left( {1,{3 \over 2}} \right)$<br><br>Equation of normal ${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$<br><br>$\Rightarrow$ ${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$<br><br>$\Rightarrow$ $4x - 2y = 1$