Mathematics - Ellipse Question with Solution | TestHub

$$x^2=8y$$ $$\frac{x^2}{3}+y^2=1$$ From (i) and (ii), $$\frac{8y}{3}+y^2=1\Rightarrow y=-3,\frac{1}{3}$$ 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 $$\left(\frac{2\sqrt{6}}{3},\frac{1}{3}\right)\text{ and }\left(-\frac{2\sqrt{6}}{3},\frac{1}{3}\right)$$ Required equation of the line, $$y-\frac{1}{3}=0\Rightarrow 3y-1=0$$