HINT:
from this and this, the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $2a(1-e^2)$ and $b^2=a^2(1-e^2)$ where $a$ is Semi major Axis, $b$ is the Semi-minor Axis and $e$ is the Eccentricity
and the length of the latus rectum of the parabola $y^2=4ax$ is $4a$
Can you utilize the relations to find the value of $e$
Check separately for cases when $a\ge b$ and when $a<b$
EDIT: after a drastic change in the question $y^2=4ax$ to $y^2=4cx$
Now WLOG, we can choose $a\ge b$
As the focus of the parabola is $(c,0)$ and those of the ellipse are $(\pm ae,0)$
$c=ae$
Now, $2a(1-e^2)=4c=4ae\implies 1-e^2=2e\implies e^2+2e-1=0$
$\implies e=\frac{-2\pm{\sqrt{2^2-4(-1)1}}}{2\cdot1}=-1\pm\sqrt2$
As $0\le e<1, e=\sqrt2-1$
$$y = x \tan \theta - \dfrac{gx^2}{2u^2 \cos^2 \theta }$$
Let $H = \dfrac{u^2 cos^2 \theta}{2g}$ and $R = \dfrac{u^2 \sin 2 \theta}{g}$
So, the given equation transforms into (by multiplying both sides with $\dfrac{2u^2 cos^2 \theta}{g}$ )
$$ 4 Hy = Rx - x^2 $$
$$\left( x - \dfrac{R}{2} \right)^2 = -4H \left(y-\dfrac{R^2}{16H}\right)$$
So, it is actually the parabola in the standard form. So, the length of latus rectum is equal to $4H$ and the vertex is $\left(\dfrac{R}{2}, \dfrac{R^2}{16H} \right)$.
Simplifying those we get,
Vertex : $\left( \dfrac{u^2 \sin 2 \theta }{2g} , \dfrac{u^2 \sin^2 \theta}{2g} \right) $
Length of Latus Rectum : $\dfrac{2u^2 cos^2 \theta}{g}$
BTW: Here $H$ is not eual to the height of projectile.
Best Answer
$$(Ax+Cy)^2+Dx+Ey+F=0\qquad\cdots (1)$$ Put $u=Ax+Cy, v=Cx-Ay$ and solving for $x,y$ and putting $K=A^2+C^2$ gives $$y=\frac {Cu-Av}{K}; \qquad x=\frac {Au+Cv}{K}\qquad\cdots (2)$$ From $(1),(2)$,
$$\begin{align} u^2+D\left(\frac {Au+Cv}{K}\right)+E\left(\frac{Cu-Av}{K}\right)+F&=0\\ \left(u+\frac {AD+CE}{2K}\right)^2&=\frac{AE-CD}{K}\bigg\lbrace v+\frac{K}{AE-CD}\left[\left(\frac{AD+CE}{2K}\right)^2-F\right]\bigg\rbrace\\ \text{Dividing by }\sqrt{K}\cdot \sqrt{K} \text{ to normalize }u,v \hspace{1cm} \\ \text{and putting }M=AE-CD,\hspace{1cm}\\ {\underbrace{\left(\frac u{\sqrt{K}}+\frac {AD+CE}{2K\sqrt{K}}\right)}_U}^2&=\underbrace{\frac{M}{K\sqrt{K}}}_{4\alpha}\underbrace{\bigg\lbrace \frac v{\sqrt{K}}+\frac{K}{M\sqrt{K}}\left[\left(\frac{AD+CE}{2K}\right)^2-F\right]\bigg\rbrace}_V\\ U^2&=4\alpha V\\ \text{Length of Latus Rectum } = |4\alpha |&=\bigg|\frac M{\;\;K^{\frac 32}}\bigg|=\color{red}{\frac {\big|AE-CD\big|}{\;\;\big(A^2+C^2\big)^{\frac 32}}} \end{align}$$