Statement 1: If the extremities of the latus rectum of the

 ellipse   $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (a> b), having positive ordinates 

lies on the parabola $x^2=-2(y-2),$ then, $a=2$

Statement 2: If the length of the latus rectum of the  

ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  is equal to the distance between 

the foci, then the eccentricity e of the ellipse satisfies 

$e^2+e-1=0$