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

The correct option is D $3y-1=0$

Graph $\frac{x^2}{3}+y^2=1$ & $x^2=8y$

Here, $x^2=8y$ is an upward facing parabola.

Let $(h,k)$ & $(-h,k)$ be the intersection point of the curves. [ we can take $(h,k)$ & $(-h,k)$ by symmetry ]

$\therefore$

$h^2=8k$ & $\frac{h^2}{3}+k^2=1$

$\Rightarrow \frac{8k}{3}+k^2=1$

$\Rightarrow 3k^2+8k-3=0$

$\therefore k=\frac{-8\pm \sqrt{64+36}}{6}=\frac{-8\pm 10}{6}$

As $k>0$, [ Because of upward facing parabola]

$k=\frac{-8+10}{6}=\frac{2}{6}=\frac{1}{3}$

$\therefore$ Equation of the line is $\Rightarrow$

$y=\frac{1}{3}\Rightarrow 3y-1=0$