Question

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$, $y_1<0,y_2<0$ be the ends of the latus rectum of the ellipse $y^2+4x^2=4$. the equations of 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. $x^2+2\sqrt{3}y=3-\sqrt{3}$
• 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 $x^2+2\sqrt{3}y=3-\sqrt{3}$

$\frac{x^2}{1}+\frac{y^2}{4}=1,$ a = 2, b = 1

the ends of latusrectum PQ are $(\pm \sqrt{a^2-b^2},-\frac{b^2}{a})$

P = $(-\sqrt{3},-\frac{1}{2})$, Q = $(\sqrt{3},-\frac{1}{2})$

The midpoint of PQ is M $(0,-\frac{1}{2})$ which is the focus of the parabola.

The vertex of the parabola is V$(0,-\frac{1}{2}\pm \frac{\sqrt{3}}{2})$

$\therefore$ The parabolas are $x^2=\pm 2\sqrt{3}(y-(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}))$

i.e., $x^2-2\sqrt{3}y=3+\sqrt{3}$ and $x^2+2\sqrt{3}y=3-\sqrt{3}$