Question

If $(u,v)$ is a point on $4x^2+a^2{y}^2=4a^2,$ where $4<a^2<8,$ that is farthest from the point $(0,-2)$ then $u+v$ is equal to

Answer 1

$4x^2+{a^{2}y}^2=4a^2$
$\frac{x^2}{a^2}+\frac{y^2}{4}=1$
Let $(u,v)$ be the point farthest from $(0,-2)$.
Let $d$ be the distance between the two points.
$\Rightarrow d^2=a^2{\cos }^2\theta +{(2\sin \theta +2)}^2$

To maximise $d^{\prime }=0$
$\Rightarrow 2d{d}^{\prime }=a^2\left(2\right(-\sin \theta \left)\cos \theta \right)+8(\sin \theta +1)\cos \theta =0$

$\cos \theta \left\{\right(8-2a^2)\sin \theta +8\}=0$

$\theta =\frac{\pi }{2}$ or $\sin \theta =\frac{4}{a^2-4}$

Now, $4<a^2<8$

$0<a^2-4<4$

$0<\frac{a^2-4}{4}<1$

$\frac{4}{a^2-4}>1$
However, $\sin \theta \le 1$

Hence, $\sin \theta =\frac{\pi }{2}$

Hence, the required point is $(0,2)$

$\Rightarrow u+v=2$