Q45 / 90 · JEE Main 2013 (09 Apr Online) · Mathematics
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 :
- A$y-3=0$
- B$y+3=0$
- C$3 y+1=0$
- Correct answerD$3 y-1=0$
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,
$$
y-\frac{1}{3}=0 \Rightarrow 3 y-1=0
$$
Hence, the correct answer is option: 'D'.