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)

$x^{2} = 8y\quad\quad\quad...\left(i\right)$
$\frac{x^{2}}{3}+y^{2} = 1\quad\quad...\left(ii\right)$
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,
$y-\frac{1}{3} = 0\quad\Rightarrow \quad3y - 1 = 0$