Given the base of a triangle and ratio of tangents of half of the base angles, Find the locus of vertex of triangle.

Question

I'm struck with a question. While finding the locus of vertex, I'm not able to eliminate the variable involved.

In the question we are given the ratio of tangents of half base angles and the length of base of triangle. We've got to find the locus of vertex of triangle (as shown in pic )

Another method of elimination could be using Quadratic formula to simply equate the roots, but this method proves to be tedious. Are there any better ways?

Here's my working- Here I've to find locus of vertex A and I've assumed the coordinates of variable point as (h,k)

(https://i.stack.imgur.com/ylaXf.jpg)

Answer 1

If $BC$ and $\frac{\tan(B/2)}{\tan(C/2)}$ are given, the projection $P$ of the incenter $I$ on $BC$ is fixed.
Let $\Gamma$ be a circle (candidate incircle) tangent to $BC$ at $P$: the tangents from $B$ and $C$ to $\Gamma$ (different from $BC$) meet on a hyperbola (can you prove it?) which is the locus of the possible $A$-vertices.

Hint(s) for proving it is a hyperbola: the involved algebra implies it is a conic, and the involved simmetries imply that such a conic is unbounded and with two branches. Blue points out a nice purely geometric alternative: $B$ and $C$ are the foci of the hyperbola above.