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(D) Given equations are:$x^2 = 8y \quad (i)$$\frac{x^2}{3} + y^2 = 1 \quad (ii)$Substituting $(i)$ into $(ii)$:$\frac{8y}{3} + y^2 = 1$$3y^2 + 8y - 3

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$3y - 1 = 0$

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(D) Given equations are:$x^2 = 8y \quad (i)$$\frac{x^2}{3} + y^2 = 1 \quad (ii)$Substituting $(i)$ into $(ii)$:$\frac{8y}{3} + y^2 = 1$$3y^2 + 8y - 3 = 0$$3y^2 + 9y - y - 3 = 0$$3y(y + 3) - 1(y + 3) = 0$$(3y - 1)(y + 3) = 0$So,$y = \frac{1}{3}$ or $y = -3$.Since $x^2 = 8y$,$y$ must be non-negative,so $y = -3$ is rejected.For $y = \frac{1}{3}$,$x^2 = 8(\frac{1}{3}) = \frac{8}{3}$,so $x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}$.The points of intersection are $(\frac{2\sqrt{6}}{3}, \frac{1}{3})$ and $(-\frac{2\sqrt{6}}{3}, \frac{1}{3})$.The line passing through these points is a horizontal line $y = \frac{1}{3}$,which simplifies to $3y - 1 = 0$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

The 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

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