locus problem in analytical geometry asking about a constant sum of two tangents to two identical circles yielding an ellipse You are given two circles:
Circle G: $(x-3)^2 + y^2 = 9$
Circle H: $(x+3)^2 + y^2 = 9$
Two lines that are tangents to the circles at point $A$ and $B$ respectively intersect at a point $P$ such that $AP + BP = 10$
Find the locus of all points $P$.


This problem is solvable if we set point $P = (x,y)$ and solve the equation $AP + BP = 10$.  After substituting  $GP^2 = AP^2 + 3^2$ and $HP^2 = BP^2 + 3^2$ and getting the following equation for an ellipse 
$16x^2 +25y^2 = 625$
That's a lot of math and algebra to do, so my question is:  What is the geometric reasoning behind why is the locus an ellipse (without using analytical geometry) or is there any other elegant proofs that lack heavy calculations?

Not answering the question but giving further observation,
\begin{align}
  \sqrt{(x-r)^2+y^2-r^2} \pm \sqrt{(x+r)^2+y^2-r^2} &= 2s \\
  \sqrt{x^2-2rx+y^2} \pm \sqrt{x^2+2rx+y^2} &= 2s \\
  2(x^2+y^2) \pm 2\sqrt{(x^2+y^2)^2-4r^2x^2} &= 4s^2 \\
  (x^2+y^2)^2-4r^2x^2 &= 4s^4-4s^2(x^2+y^2)+(x^2+y^2)^2 \\
  (s^2-r^2)x^2+s^2y^2 &= s^4 \\
\end{align}


*

*Positive sign is taken when $s^2>r^2$ giving an ellipse.

*Negative sign is taken for constant difference instead, the locus can be two horizontal lines $(s^2=r^2)$, a hyperbola $(s^2<r^2)$ or a vertical line $(s=0)$.

*The loci always pass through the point $(0,\pm s)$.

*The loci don't have contact with two circles when $s^2>2r^2$.

*The eccentricity is given by $e=\dfrac{r}{|s|}$.
