QJEEAdv08Ma66df: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma66df Solution: Let and be the end points of the latus rectum of the ellipse…

The ellipse is ${x^2} + 4{y^2} = 4$ ,

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

We have ${b^2} - {a^2}(1 - {e^2})$

$e = {{\sqrt 3 } \over 2}$

P and Q are obtained as

$P\left( {\sqrt 3 ,{{ - 1} \over 2}} \right),Q = \left( { - \sqrt 3 ,{{ - 1} \over 2}} \right)$

We have $PQ = 2\sqrt 3$

With PQ as latus rectum two parabolas are possible their vertices being

$\left( {0,{{ - \sqrt 3 - 1} \over 2}} \right)$ and $\left( {0,{{\sqrt 3 - 1} \over 2}} \right)$

The equation of the parabola s can be obtained as

${x^2} - 2\sqrt 3 y = 3 + \sqrt 3$ and ${x^2} + 2\sqrt 3 y = 3 - \sqrt 3$