Finding eccentricity of an ellipse from latus rectum The latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the same as latus rectum of a parabola $y^2=4cx$ . Find eccentricity of the ellipse .

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 $0\le e<1, e=\sqrt2-1$
