Coordinates of the vertices $B$ and $C$of a triangle $ABC$are $(2,0)$ and $(8,0)$respectively. The vertex $A$is varying in such a way that $4\tan \frac{B}{2}\tan \frac{C}{2}=1$and locus of $A$is $\frac{{(x-5)}^2}{25}+\frac{y^2}{k^2}=1$, then $k=$

$4\tan \frac{B}{2}\tan \frac{C}{2}=1\Rightarrow \sqrt{\frac{(s-c)(s-a)}{s(s-b)}\times \frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{4}$

$\Rightarrow \frac{s-a}{s}=\frac{1}{4}\Rightarrow \frac{2s-a}{a}=\frac{5}{3}$

$\Rightarrow b+c=\frac{5}{3}\times 6=10[\because a=BC=6]$

Thus, sum of distance of variable point $A$from two given fixed points $B$and $C$is always 10. Therefore, equation of locus of $A$is $\frac{{(x-5)}^2}{25}+\frac{y^2}{16}=1$