A triangle in ellipse Coordinates of the vertices B and C of a triangle ABC are (2,0) and (8,0)respectively.The vertex A is varing in such a way that $4\tan(B/2)\tan(C/2)=1$.then the locus of A has to be find.
Now , i didnt get any idea about the same.
Pls help me solving this.

Applying Componendo-Dividendo we get $\dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)} = \dfrac{3}{5}$
We can easily prove that $\dfrac{a}{b+c} = \dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)}$
Thus $b+c = 10$ and hence locus of $A$ is an ellipse with focii at $(2,0)$ and $(8,0)$ and major axis $=10$. Then the center is at $(5,0)$ and we get semi-minor axis is $4$.
So the equation of the ellipse is $\dfrac{(x-5)^2}{25}+\dfrac{y^2}{16}=1$
