Question

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ where $y_1,y_2<0,$ be the end points of the latus rectum of the ellipse $x^2+4y^2=4.$ Then equation(s) of the parabola with latus rectum $PQ$ is/are

• 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. $x^2-2\sqrt{3}y=3-\sqrt{3}$

Answer 1

Answer: (B), (C)

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


The given ellipse is
$\frac{x^2}{4}+\frac{y^2}{1}=1$
$\Rightarrow e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

Hence, the end points $P$ and $Q$ of the latus rectum are given by
$P\equiv (\sqrt{3},-\frac{1}{2})$ and
$Q\equiv (-\sqrt{3},-\frac{1}{2})$ (given $y_1,y_2<0$)

Coordinates of midpoint of $PQ$ are
$R\equiv (0,-\frac{1}{2})$
Length of latus rectum $PQ=2\sqrt{3}$

Hence, two parabolas are possible whose vertices are
$(0,-\frac{1}{2}-\frac{\sqrt{3}}{2})$
and
$(0,-\frac{1}{2}+\frac{\sqrt{3}}{2})$ be on

So the equation(s) of the parabola are
$(x-0{)}^2=2\sqrt{3}(y+\frac{1}{2}+\frac{\sqrt{3}}{2})$ and $(x-0{)}^2=-2\sqrt{3}(y+\frac{1}{2}-\frac{\sqrt{3}}{2})$
$\Rightarrow x^2-2\sqrt{3}y=3+\sqrt{3}$ and $x^2+2\sqrt{3}y=3-\sqrt{3}$