Question

$P(x_1,y_1)$ and $Q(x_2,y_2)$, $y_1<0$, $y_2<0$ are the end points of the latus rectum of the ellipse $x^2+4y^2=4$. The equations of the parabolas with latus rectum $PQ$ are

• A. $x^2+2\sqrt{3}y=3+\sqrt{3}$
• B. $x^2-2\sqrt{3}y=3+\sqrt{3}$
• C. $x^2+2\sqrt{3}y=\sqrt{3}-3$
• D. $x^2-2\sqrt{3}y=3-\sqrt{3}$

Answer 1

Answer: (B), (C)

The correct options are
B $x^2-2\sqrt{3}y=3+\sqrt{3}$
C $x^2+2\sqrt{3}y=\sqrt{3}-3$

Eccentricity e of the ellipse is given by
$b^2=a^2(1-e^2)$
$\Rightarrow$ $1=4(1-e^2)\Rightarrow e=\frac{\sqrt{3}}{2}$.
Focii of the ellipse are $(\sqrt{3},0)$ and $(-\sqrt{3},0)$.
Length of a latus rectum of the ellipse is
$2\frac{b^2}{a}=1$
Thus, $P(x_1,y_1)=P(-\sqrt{3},-\frac{1}{2})$
and $Q(x_2,y_2)=Q(\sqrt{3},-\frac{1}{2})$
Length of the latus rectum $PQ$ of the parabola is
$|x_2-x_1|=2\sqrt{3}=4p$(say)
As the focus of a parabola is the midpoint of the latus rectum, focus of the desired parabola is $(0,-\frac{1}{2})$ and hence its vertices are $(0,-\frac{1}{2}\pm p)$
i.e. $(0,-\frac{1}{2}-\frac{\sqrt{3}}{2})$ and $(0,-\frac{1}{2}+\frac{\sqrt{3}}{2})$
Thus, there are two parabolas having PQ as the latus rectum whose equations are
$x^2=4p(y+\frac{1}{2}+\frac{\sqrt{3}}{2})=2\sqrt{3}y+\sqrt{3}+3$
and $x^2=-4p(y-\frac{1}{2}-\frac{\sqrt{3}}{2})=-2\sqrt{3}y+\sqrt{3}-3$