Question

The base of a triangle and the ratio of the tangents of half angle on the base are given. Prove that the locus of the vertex of the triangle is a hyperbola.

Answer 1

Let $BC$ be the base of the triangle and $A$ is the vertex.

$\frac{\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{C}{2}\right)}=\frac{\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}{\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}}=\frac{s-c}{s-b}=\frac{2s-2c}{2s-2b}=\frac{a+b-c}{a+c-b}=k$ (let it be.)

By componendo-diviendo, we have

$\Rightarrow \frac{k-1}{k+1}=\frac{b-c}{a}$

$\Rightarrow b-c=a\left(\frac{k-1}{k+1}\right)$ =constant

$\Rightarrow AC-AB=constant$

Therefore, the difference of distance of point $A$ from the two foci $B$ and $C$ is constant.

So, the locus of point $A$ is a $\text{hyperbola}$.