EllipseHardBloom L3

Question

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 the latus rectum of ellipse $x^{2} + 4y^{2} = 4$. The equations of parabolas with latus rectum $PQ$ are:

Options

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

Sol.

$$1 = 4\left( 1 - e^{2} \right) \Rightarrow e = \frac{\sqrt{3}}{2}$$

Lengths of latus rectum of parabola $= 4a = 2\sqrt{3}$

$$\Rightarrow a = \frac{\sqrt{3}}{2} $$Equation of parabolas are

$$2\sqrt{3}\left( y - \left( - \frac{1}{2} - \frac{\sqrt{3}}{2} \right) \right) = x^{2}$$

or $- 2\sqrt{3}\left( y - \left( - \frac{1}{2} + \frac{\sqrt{3}}{2} \right) \right) = x^{2}$

$$\begin{matrix} & \ 2\sqrt{3}y + (1 + \sqrt{3})\sqrt{3} = x^{2} \\ & \text{~}\text{or}\text{~}\ - 2\sqrt{3}y - (1 - \sqrt{3})\sqrt{3} = x^{2} \\ & \ x^{2} - 2\sqrt{3}y = \sqrt{3} + 3 \\ & \text{~}\text{or}\text{~}\ x^{2} + 2\sqrt{3}y = 3 - \sqrt{3} \end{matrix}$$

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