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 :

To find the equation of the line passing through the points of intersection of the parabola $$x^2 = 8y$$ and the ellipse $$\frac{x^2}{3} + y^2 = 1$$, we need to solve these equations simultaneously. The points where both curves intersect will satisfy both equations, and the line joining these points can be found.

First, write the equations clearly:

Parabola: $$x^2 - 8y = 0$$ ...(1)

Ellipse: $$\frac{x^2}{3} + y^2 - 1 = 0$$ ...(2)

From equation (1), express $$x^2$$ in terms of $$y$$:

$$x^2 = 8y$$

Substitute this into equation (2) to eliminate $$x^2$$:

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

Multiply both sides by 3 to clear the denominator:

$$8y + 3y^2 = 3$$

Bring all terms to one side to form a quadratic equation:

$$3y^2 + 8y - 3 = 0$$

Solve this quadratic equation for $$y$$. The discriminant $$D$$ is given by $$D = b^2 - 4ac$$, where $$a = 3$$, $$b = 8$$, and $$c = -3$$:

$$D = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100$$

Now, find the roots:

$$y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{100}}{6} = \frac{-8 \pm 10}{6}$$

So, the two solutions are:

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

$$y = \frac{-8 - 10}{6} = \frac{-18}{6} = -3$$

For each $$y$$, find the corresponding $$x$$ using $$x^2 = 8y$$ from equation (1).

First, for $$y = \frac{1}{3}$$:

$$x^2 = 8 \cdot \frac{1}{3} = \frac{8}{3}$$

$$x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}$$

Second, for $$y = -3$$:

$$x^2 = 8 \cdot (-3) = -24$$

Since $$x^2$$ cannot be negative for real $$x$$, there are no real solutions for $$y = -3$$.

Thus, the only real points of intersection are $$\left( \frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$ and $$\left( -\frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$. Both points have the same $$y$$-coordinate, $$y = \frac{1}{3}$$.

The line joining these points is horizontal, given by $$y = \frac{1}{3}$$. Rewriting this in standard form:

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

Multiply both sides by 3 to clear the fraction:

$$3y - 1 = 0$$

Now, verify that these points satisfy both original equations.

For the point $$\left( \frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$:

Parabola: $$x^2 = \left( \frac{2\sqrt{6}}{3} \right)^2 = \frac{4 \cdot 6}{9} = \frac{24}{9} = \frac{8}{3}$$ and $$8y = 8 \cdot \frac{1}{3} = \frac{8}{3}$$, so equal.

Ellipse: $$\frac{x^2}{3} + y^2 = \frac{\frac{8}{3}}{3} + \left( \frac{1}{3} \right)^2 = \frac{8}{9} + \frac{1}{9} = \frac{9}{9} = 1$$, so satisfied.

The same holds for $$\left( -\frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$ due to symmetry.

Comparing with the options:

A. $$y - 3 = 0$$

B. $$y + 3 = 0$$

C. $$3y + 1 = 0$$

D. $$3y - 1 = 0$$

The equation $$3y - 1 = 0$$ matches our result.

Hence, the correct answer is Option D.