Let $$P$$ be a parabola with vertex $$(2, 3)$$ and directrix $$2x + y = 6$$. Let an ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$ of eccentricity $$\frac{1}{\sqrt{2}}$$ pass through the focus of the parabola $$P$$. Then the square of the length of the latus rectum of $$E$$, is

Let the given parabola $$P$$ have vertex $$V(2,3)$$ and directrix $$L:2x+y-6=0$$.
Distance from $$V$$ to $$L$$ is $$ p \;=\;\frac{\bigl|2\cdot 2 + 3 -6\bigr|}{\sqrt{2^2+1^2}} \;=\;\frac{1}{\sqrt{5}}. $$

The axis of the parabola is perpendicular to the directrix, so its direction is the unit normal to $$L$$, $$ \mathbf{n}=\frac{(2,1)}{\sqrt{5}}. $$ 

Since the vertex lies in the half-plane $$2x+y-6>0$$, the focus lies in the same half-plane. 

Thus the focus is $$ F \;=\;V + p\,\mathbf{n} =\;(2,3)+\frac{1}{\sqrt{5}}\;\frac{(2,1)}{\sqrt{5}} =\Bigl(2+\tfrac{2}{5},\,3+\tfrac{1}{5}\Bigr) =\Bigl(\tfrac{12}{5},\,\tfrac{16}{5}\Bigr). $$

Next, let the ellipse $$E$$ be $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b, \quad\text{with eccentricity }e=\frac{1}{\sqrt{2}}. $$ 

We have the standard formula for an ellipse, $$ e=\frac{c}{a},\quad c^2=a^2-b^2. $$ Since $$e^2=\tfrac12$$, it follows $$ c^2=a^2\,e^2=\frac{a^2}{2}, \quad b^2=a^2-c^2=a^2-\frac{a^2}{2}=\frac{a^2}{2}. $$

 $$ \frac{x^2}{a^2}+\frac{y^2}{\tfrac{a^2}{2}}=1 \quad\Longrightarrow\quad \frac{x^2}{a^2}+\frac{2y^2}{a^2}=1 \quad\Longrightarrow\quad x^2+2y^2=a^2. $$

Since the focus $$F\bigl(\tfrac{12}{5},\tfrac{16}{5}\bigr)$$ of the parabola lies on this ellipse, substitute into $$x^2+2y^2=a^2$$: $$ \Bigl(\tfrac{12}{5}\Bigr)^2+2\Bigl(\tfrac{16}{5}\Bigr)^2 =\frac{144}{25}+\frac{512}{25} =\frac{656}{25} \;=\;a^2. $$

The length of the latus rectum of the ellipse is given by $$ \ell=\frac{2b^2}{a} =\frac{2\bigl(\tfrac{a^2}{2}\bigr)}{a} =\frac{a^2}{a} =a, $$ so the square of its length is $$ \ell^2=a^2=\frac{656}{25}. $$