The eccentricity of an ellipse whose centre is at the origin is \[{1 \over 2}\]. If one of its directrices is x = -- 4, then the
equation of the normal to it at \[\left( {1,{3 \over 2}} \right)\] is :
Solutions · 1
Official solution
Given e = \[{1 \over 2}\] and \[{a \over e}\] = 4
\[ \therefore \] a = 2
We have b2 = a2 (1 -- e2) = \[4\left( {1 - {1 \over 4}} \right)\] = 3
\[ \therefore \] Equation of ellipse is
\[{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1\]
Now, the equation of normal at \[\left( {1,{3 \over 2}} \right)\] is
\[{{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}\]
\[ \Rightarrow \] \[{{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3\]
\[ \Rightarrow \] 4x -- 2y = 1