Step 1: Address the apparent errors in the question statement.
The question is poorly phrased. It mentions an "angle of intersection $\theta$" and "the point 20/3 on the ellipse", neither of which seems relevant to finding the final answer, which is expressed in terms of 'a'. A plausible interpretation is that the question is simply asking for the coordinates of one of the intersection points of the two conics. We will proceed under this assumption.
Step 2: Use the given eccentricity to find the equation of the ellipse.
The ellipse is a vertical ellipse since $b>a$. The eccentricity formula is $e^2 = 1 - \frac{a^2}{b^2}$.
We are given $e = \frac{1}{\sqrt{2}}$, so $e^2 = \frac{1}{2}$.
\[
\frac{1}{2} = 1 - \frac{a^2}{b^2} \implies \frac{a^2}{b^2} = \frac{1}{2} \implies b^2 = 2a^2.
\]
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{2a^2} = 1$, which can be rewritten as $2x^2+y^2=2a^2$.
Step 3: Find the intersection points of the ellipse and the parabola $y^2=4ax$.
We substitute $y^2=4ax$ from the parabola's equation into the ellipse's equation.
\[
2x^2 + (4ax) = 2a^2.
\]
\[
2x^2 + 4ax - 2a^2 = 0 \implies x^2 + 2ax - a^2 = 0.
\]
Solving for $x$ using the quadratic formula:
\[
x = \frac{-2a \pm \sqrt{(2a)^2 - 4(1)(-a^2)}}{2} = \frac{-2a \pm \sqrt{8a^2}}{2} = \frac{-2a \pm 2a\sqrt{2}}{2} = a(-1 \pm \sqrt{2}).
\]
Since $y^2=4ax$ and we assume $a>0$, we must have $x \ge 0$. So we take the positive root: $x = a(\sqrt{2}-1)$.
The corresponding y-coordinate is $y^2 = 4a(a(\sqrt{2}-1)) = 4a^2(\sqrt{2}-1)$, so $y = \pm 2a\sqrt{\sqrt{2}-1}$.
These intersection points do not match any of the options.
Step 4: Re-evaluate the question and check the options.
Given the inconsistency, it's highly likely the problem statement has errors. Let's test if the point from the keyed answer, (D) $(\frac{a}{\sqrt{2}}, \frac{\sqrt{3}a}{\sqrt{2}})$, lies on either curve.
Check on the ellipse $2x^2+y^2=2a^2$:
\[
2\left(\frac{a}{\sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}a}{\sqrt{2}}\right)^2 = 2\left(\frac{a^2}{2}\right) + \frac{3a^2}{2} = a^2 + \frac{3a^2}{2} = \frac{5a^2}{2}.
\]
Since $\frac{5a^2}{2} \neq 2a^2$, the point is not on the ellipse.
Check on the parabola $y^2=4ax$:
\[
\left(\frac{\sqrt{3}a}{\sqrt{2}}\right)^2 = \frac{3a^2}{2}. \quad \text{and} \quad 4a\left(\frac{a}{\sqrt{2}}\right) = 2\sqrt{2}a^2.
\]
Since $\frac{3a^2}{2} \neq 2\sqrt{2}a^2$, the point is not on the parabola either.
The question is fundamentally flawed. To arrive at the keyed answer (D), one would need a different set of initial equations.