If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4<a^2<8$,that is farthest from $(0,-2)$ then $u+v$ is equal to?
My Approach:
I took a parametric point $(t,4-4t^2/a^2)$.And then tried to find the minima of the distance.But that is too lengthy method.Any other suggestions?

HINT:
WLOG any point on the ellipse $$(a\cos t,2\sin t)$$
So, if the distance is $d,$ 
$$d^2=(a\cos t)^2+(2+2\sin t)^2=8+4\sin t+(a^2-4)\cos^2t$$
$$=8+\sqrt{4^2+(a^2-4)^2}\sin\left(t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}\right)$$
$$\le8+\sqrt{4^2+(a^2-4)^2}$$
Now $4<a^2<8,0<a^2-4<4\implies(a^2-4)^2\le4^2$
So, the maximum value i.e., the equality occurs if $t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}=2m\pi+\dfrac\pi2\iff t=2m\pi+\arcsin\dfrac4{\sqrt{4^2+(a^2-4)^2}}$
and $a^2-8$ is maximum which can be made arbitrarily small but will remain negative.
