Projectile motion: Proving:$ x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2} $ 

Question: Projectiles are fired with initial speed $v$ and variable launch angle $0< \alpha < \pi$.
Choose a coordinate system with the firing position at the origin. For each value of $\alpha$ the trajectory will follow a parabolic arc with apex at $(x,y)$ where both $x$ and $y$ depend on $\alpha$. Show that:
$$ x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2 = \frac{v^2}{4g^2} $$
and hence the points of maximum heigh of the trajectories lie on an ellipse.
Requires some basic understanding of physics.



What I have attempted: 
As the initial velocity is $v$ , resolving the components 
$$ v_x = v\cos(\alpha) $$
$$ v_y = v\sin(\alpha) $$
To find the time to apex (vertex) at $(x,y)$ would be using the formula 
$$ v_f = v_i + at $$
$v_f$ at the top is 0 and $a=-g$ and $v_i =  v\sin(\alpha)$ hence
$$ t_{top} = \frac{ v\sin(\alpha)}{g} $$
and by using the formula $$ y = v_yt + \frac{1}{2}a_yt^2 $$
$$ y = \frac{1}{2}g t^2 $$
$$ y_{apex} = \frac{v^2\sin^2(\alpha)}{2g} $$
And as horizontal movement is at constant velocity
$$  x = v_x \cdot t $$
$$ x= v\cos(\alpha) \cdot \frac{ v\sin(\alpha)}{g} $$
$$ x = \frac{v^2 \sin(\alpha)\cos(\alpha)}{g} $$
$$ x_{apex}  = \frac{v^2 \sin(2\alpha)}{2g} $$
Now I am stuck how do I prove the equation.... 

We have:
$$4\left(y-\frac{v^2}{4g}\right)^2=4\left(\frac{v^2\sin^2(\alpha)}{2g}-\frac{v^2}{4g}\right)^2=\frac{v^4\sin^4(\alpha)}{g^2}-\frac{v^4\sin^2(\alpha)}{g^2}+\frac{v^4}{4g^2}$$
$$=\left(\frac{v^4\sin^2\alpha}{g^2}\right)(\sin^2\alpha-1)+\frac{v^4}{4g^2}$$
$$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)(1-\sin^2\alpha)+\frac{v^4}{4g^2}$$
$$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)\cos^2\alpha+\frac{v^4}{4g^2}$$
and
$$x^2= \frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}$$
so 
$$x^2 + 4 \left(y-\frac{v^2}{4g} \right)^2=\frac{v^4 \sin^2\alpha\cos^2\alpha}{g^2}-\frac{v^4\sin^2\alpha\cos^2\alpha}{g^2}+\frac{v^4}{4g^2}$$
$$=\frac{v^4}{4g^2}$$
