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.
 A: Let $I$ be the incenter and let the incircle touch $BC$ at $D$. Then $BD = s-b$, $CD = s-c$ and $DI =r$ (inradius). So if you look at triangle $BDI$ and $CDI$ we have
$$\tan B/2 = {r\over s-b}\;\;\;{\rm and}\;\;\; \tan C/2 = {r\over s-c}$$ where $r=A/s$ and $s$ is semiperimeter of $ABC$. Then we have $$4r^2 = (s-b)(s-c)$$ and $$r^2 = {A^2\over s^2} = {(s-a)(s-b)(s-c)\over s}$$ thus $$ 4(s-a) =s\;\;\;\Longrightarrow \;\;\;3(b+c) = 5a = 30$$ so $b+c = 10$. Thus $A$ describe ellipse with focuses at $B$ and $C$. 
Explicit formula for this ellipse is $${(x-5)^2\over 25}+{y^2\over 16}=1$$ 
A: 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 $\dfrac{(x-5)^2}{25}+\dfrac{y^2}{16}=1$
