Distance between the curves from a particular point The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points  $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to
(A) 1
(B) 2
(C) 3
(4) 4
My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.
The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$
$D^2=(u-2)^2+v^2$
Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$
How we will proceed from here

Distance $D_1$ of P(u,v) from the circle is $D_1=\sqrt{(u+2)^2+v^2}-2=\sqrt{F}-2$
and $D_2=\sqrt{(u-2)+v^2}=\sqrt{G}$, we require
$$D_1=D_2 \implies \sqrt{(u+2)^2+v^2}-2=\sqrt{(u-2)+v^2} \implies \sqrt{F}-\sqrt{G}=2~~~(1)$$ We also have $$F-G=8u \implies \sqrt{F}+\sqrt{G}=4u~~~~(2)$$
From (1) and (2), we have $$\sqrt{F}=1+2u \implies u^2+4u+4+qu^2-q=1+4u+4u^2 \implies (q-3) u^2=(q-3)\implies q=3$$
