Question

Given the base of a triangle and the ratio of the tangent of half the base angles,find the locus of vertex.

Answer 1

Suppose $AB$ to the base and $C$ be the moving point,

Then $A$ and $B$ are the base angle and by hypothesis is

$\frac{\tan \left(\frac{A}{2}\right)}{\tan \left(\frac{B}{2}\right)}=\frac{\frac{r}{(s-a)}}{\frac{r}{(s-b)}}=\frac{s-b}{s-a}=$ constant $=\lambda$

$\Rightarrow \lambda =\frac{\frac{a+b+c}{2}-a}{\frac{a+b+c}{2}-b}=\frac{b+c-a}{c+a-b}$

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

Hence, by componendo and dividendo

$\frac{1+\lambda }{1-\lambda }=\frac{2c}{2(b-a)}=\frac{c}{b-a}\Rightarrow b-a=c\frac{(1-\lambda )}{(1+\lambda )}$

As $c$ is a constant hence, $(b-a)=$ difference of distances of a point from two fixed points$=$ constant

Therefore, the locus of the vertex $C$ is a hyperbola.