The co-ordinates of the focii are $(h\pm ae)$, so your $c$ will be $ae$ rather than $\frac{e}{a}$ and so your $a$ will be $\sqrt2$ and your $b$ will be $\sqrt\frac{3}{2}$.
So the equation we have in the rotated system is:$$\frac{(x-\frac{1}{\sqrt2})^2}{2}+\frac{2y^2}{3}=1$$
Now all that remains is to somehow rotate this back into the original system.
Here is a purely analytical solution. Canonical parabola equation is
$$
y^2=2px
$$
with focus in $(p/2,0)$. The tangent line to point $(x_0,y_0)$ is
$$
px - y_0 y+px_0=0
$$
That is all we need. Tangent line directional vector is $\vec{u}=\{y_0,p\}$. Vector $\vec{v}=\{p,-y_0\}$ is orthogonal to $\vec{u}$ and $u=|\vec{u}|=|\vec{v}|$. Consider vector $\vec{q}=\{p/2-x_0,0\}$ directional for the line coming through the focus $(p/2,0)$ and point $(x_0,y_0)$ belonging to the parabola. Its decomposition in directions of $\vec{u}$ and $\vec{v}$ looks as follows:
$$
\vec{q}=\frac{\vec{q}\cdot\vec{u}}{uq}\vec{u}+\frac{\vec{q}\cdot\vec{v}}{uq}\vec{v}
$$
Now the important part: directional vector $\vec{q}'$ of the reflection would have $\vec{u}$-component opposite to that of $\vec{q}$ and $\vec{v}$-component the same. Thus,
$$
\vec{q}'=-\frac{\vec{q}\cdot\vec{u}}{uq}\vec{u}+\frac{\vec{q}\cdot\vec{v}}{uq}\vec{v}
$$
All it's left is to calculate the relevant dot-products and to see that $y$-component of $\vec{q}'$ is zero. The dot-products:
$$
\vec{q}\cdot\vec{u}=(p/2-x_0)y_0-y_0p=-y_0(x_0+p/2)
$$
and
$$
\vec{q}\cdot\vec{v}=(p/2-x_0)p+y_0^2=p/2-px_0+2px_0=p(x_0+p/2)
$$
Finally, $y$-component of $\vec{q}'$ is
$$
\frac{1}{qu}\left[y_0(x_0+p/2)p+p(x_0+p/2)(-y_0)\right]=0
$$
QED
Best Answer
Figure 1: Start with a line $l$ and and two arbitrary points $A$ and $B$ on the same side of the line.
Figure 2: Suppose line $l$ is actually a mirror. By Fermat's principle we know that light always follows the shortest path. Therefore of all the possible paths from $A$ to some arbitrary point $X$ along $l$, to $B$, the one that light will follow is that which minimizes distance $AXB$
Figure 3: To find the path minimizing $AXB$, reflect point $A$ about line $l$ (call this reflection $A'$). Note that by construction $\triangle AXA'$ is an isosceles triangle. It follows that distance $AXB$ equals distance $A'XB$. Since this is true, and since the shortest distance between two points is a straight line, distance $AXB$ is minimized when point $X$ takes the position along line $l$ such that $A'XB$ is a straight line. This position is indicated in figure 3 by point $C$.
Since $\triangle ACA'$ is by construction isosceles, it is not hard to see that (by the properties of isosceles triangles and interior opposite angles) the angles indicated in figure 3 are equal. That is, the angle of the light ray's incidence equals the angle of its reflection off the mirror. This what is referred to as the reflective or optical or specular property of light rays.
Figure 4: Finally, to see that the ellipse also has the reflective property, suppose you are told that line $l$ is tangent to an ellipse with focii $A$ and $B$, but you are not given the point where $l$ actually touches the ellipse. To find this point of tangency the process is the same as for finding the distance-minimizing path of light. Clearly (by the definition of tangents), for any arbitrary point $X$ along $l$, with $X \neq C$, the inequality $AXB>ACB$ holds. Thus it is evident that the distance $AXB$ will be minimized at the point of tangency $C$. We have just seen that this is exactly the optical or reflective property of light. Therefore the indicated angles are equal and any ray shot from one focus to the ellipse perimeter will end up bouncing to the other focus.