Suppose you have the focus at point $(a, b)$ and a directrix with equation $cx+dy=e$. Then a point $(x,y)$ is on the parabola if it has the same distance from focus and directrix, which means
$$(a-x)^2+(b-y)^2=\frac{(cx+dy-e)^2}{c^2+d^2}$$
You can rewrite that as
$$
(x,y,1)
\begin{pmatrix}
d^2 & -cd & ce-a(c^2+d^2) \\
-cd & c^2 & de-b(c^2+d^2) \\
ce-a(c^2+d^2) & de-b(c^2+d^2) & (a^2+b^2)(c^2+d^2) - e^2
\end{pmatrix}
\begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0
$$
If you choose the Hesse normal form for the line, with unit length normal vector, you have $c^2+d^2=1$ which will make the matrix easier to read. Note that the entries of this matrix correspond to the coefficients Awesome assumed in his comment.
Do this for both your parabolas, and you get two matrices which you can then feed into the generic method for intersecting conics.
I'm assuming there is some educational reason why you're not supposed to use
the formula $d_1+d_2=2a$.
That formula gives the correct result for good reasons.
The method in which you solve two simultaneous quadratic equations
is legitimate; after eliminating $y$, we can show that $x(x-2)=0$.
If we allow complex solutions, this leads to four solutions;
but since the points on the ellipse have real coordinates,
it is perfectly legitimate to discard the complex solutions.
The fact that $x$ and $y$ must be real is simply a constraint on the
solution space; you can even add it to your system of equations as follows:
\begin{gather}
\tfrac94(x-x_0)^2 + 3y^2 = 1,\\
x^2 + y^2 = 1,\\
\Im (x) = 0,\\
\Im (y) = 0.
\end{gather}
The last two equations simply say that $x$ and $y$ have zero imaginary parts.
Another technique is to convert the equation of the ellipse to
polar coordinates. The ellipse equation you were given was in the
general form
$$
\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1
$$
for positive $a$ and $b$,
where in this particular case $x_0=\frac13$, $y_0=0$, $a=\frac23$,
and $b=\frac{1}{\sqrt3}$.
Since $\frac23 > \frac{1}{\sqrt3}$, the length of the semi-major axis is $a$.
In polar coordinates, the general equation for an ellipse with
semi-major axis $a$, semi-minor axis $b$,
one focus at $(0,0)$, and the other focus on the positive $x$-axis,
like this ellipse, is
$$
r = \frac{a(1 - e^2)}{1 - e\cos\theta}
$$
where $e = \sqrt{1 - \left(\frac ba\right)^2}$.
Plugging in $a=\frac23$ and $b=\frac{1}{\sqrt3}$, we find that
$e = \sqrt{1 - \frac34} = \frac12$, so the equation of this
particular ellipse comes out to
$$
r = \frac{\frac23\left(1 - \left(\frac12\right)^2\right)}
{1 - \frac12\cos\theta}
= \frac{1}{2 - \cos\theta}.
$$
But since $r$ in this equation is simply the distance from $(0,0)$
to a point on the ellipse at angle $\theta$ counterclockwise from
the positive $x$-axis, and the desired point $P$ is at distance
$\frac12$ from $(0,0)$, the polar coordinates of $P$ must satisfy
$$
\frac12 = \frac{1}{2 - \cos\theta},
$$
from which we can deduce that $\cos\theta = 0$,
and therefore $\sin\theta = \pm 1$.
Converting the polar coordinates back into $x,y$ coordinates gives us
\begin{align}
x &= r \cos\theta = 0,\\
y &= r \sin\theta = \pm\frac12.
\end{align}
Therefore $P = \left(0,\frac12\right)$ or $P = \left(0,-\frac12\right)$.
Best Answer
Put the ellipse in standard form. Start from $4x^2+24y^2=65$. Divide through by $65$. We get $\frac{4}{65}x^2+\frac{24}{65}y^2=1$.
This is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=\frac{65}{4}$ and $b^2=\frac{65}{24}$. Finally, you want $\sqrt{a^2-b^2}$.
Your intuition is right: this is substantially smaller than your answer. To two decimal places, I get $3.68$.