Equation of the line passing through the points of intersection of the parabola $x^2=8 y$ and the ellipse…
Equation of the line passing through the points of intersection of the parabola $x^2=8 y$ and the ellipse $\frac{x^2}{3}+y^2=1$ is :
$y-3=0$
$y+3=0$
$3 y+1=0$
$3 y-1=0$
Solution
$
x^2=8 y
$
$
\frac{x^2}{3}+y^2=1
$
From (i) and (ii),
$
\frac{8 y}{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) \text { and }\left(-\frac{2 \sqrt{6}}{3}, \frac{1}{3}\right)
$
Required equation of the line,
$
y-\frac{1}{3}=0 \Rightarrow 3 y-1=0
$