JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 18)
Let $$P\left( {{x_1},{y_1}} \right)$$ and $$Q\left( {{x_2},{y_2}} \right),{y_1} < 0,{y_2} < 0,$$ be the end points of the latus rectum of the ellipse $${x^2} + 4{y^2} = 4.$$ The equations of parabolas with latus rectum $$PQ$$ are :
$${x^2} + 2\sqrt 3y = 3 + \sqrt 3 $$
$${x^2} - 2\sqrt 3y = 3 + \sqrt 3 $$
$${x^2} + 2\sqrt 3y = 3 - \sqrt 3 $$
$${x^2} - 2\sqrt 3 y = 3 - \sqrt 3 $$
Explanation
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 $${x^2} + 2\sqrt 3 y = 3 - \sqrt 3 $$
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