Question

Equation of the line passing through the points of intersection of the parabola $x^2=8y$ and the ellipse $\frac{x^2}{3}+y^2=1$ is :

• A. $y-3=0$
• B. $y+3=0$
• C. $3y+1=0$
• D. $3y-1=0$

Answer 1

Answer: (D)

$x^2=8y...\left(i\right)$
$\frac{x^2}{3}+y^2=1...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$\frac{8y}{3}+y^2=1\Rightarrow y=-3,\frac{1}{3}$
When $y=-3$, then $x^2=-24$, which is not possible.
When $y=\frac{1}{3}$, then $x=\pm \frac{2\sqrt{6}}{3}$
Point of intersection are
$\left(\frac{2\sqrt{6}}{3},\frac{1}{3}\right)and\left(-\frac{2\sqrt{6}}{3},\frac{1}{3}\right)$
Required equation of the line,
$y-\frac{1}{3}=0\Rightarrow 3y-1=0$