Equation of the line passing through the points of intersection of...

x²=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