First slide

Ellipse

Question

$Let P (x_{1}, y_{1}) and Q (x_{2}, y_{2}), y_{1} < 0, y_{2} < 0,$ be the endpoints of the latus rectum of the ellipse x² + 4y² = 4. The equations of parabolas with latus rectum PQ are

Moderate

A

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

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

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

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

Solution

The given ellipse is
$\begin{array}{l} \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1 \\ b^{2} = a^{2} (1 - e^{2}) \\ or e = \frac{\sqrt{3}}{2} \end{array}$
Hence, the endpoints P and Q of the latus rectum are given by
$\begin{array}{l} P \equiv (\sqrt{3}, - \frac{1}{2}) \\ and Q \equiv (- \sqrt{3}, - \frac{1}{2}) (Given y_{1}, y_{2} < 0) \end{array}$
The midpoint of PQ is
$R \equiv (0, - \frac{1}{2})$
length of latus rectum, $PQ = 2 \sqrt{3}$
Hence, two parabolas are possible whose vertices are
$S (0, + \frac{\sqrt{3}}{2} - \frac{1}{2}) and T (0, - \frac{\sqrt{3}}{2} - \frac{1}{2})$
the equations of the Parabolas are
$\begin{array}{ll} x^{2} = 2 \sqrt{3} (y + \frac{\sqrt{3}}{2} + \frac{1}{2}) \\ and x^{2} = - 2 \sqrt{3} (y - \frac{\sqrt{3}}{2} + \frac{1}{2}) \\ or x^{2} - 2 \sqrt{3} y = 3 + \sqrt{3} \\ and x^{2} + 2 \sqrt{3} y = 3 - \sqrt{3} \end{array}$

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