Question

The base $BC$ $(=2a)$ of a triangle $ABC$ is fixed; the axes being $BC$ and a perpendicular to it through its middle point, find the locus of the vertex $A$, when the tangent of one base angle is $m$ times the tangent of the other.

Answer 1

Let $O$ be the mid point of base $BC$ of triangle $ABC$ and the vertex $A$ be $(h,k)$

In $△ABD\tan x=\frac{k}{h+a}....\left(i\right)$

In $△ACD\tan \varphi =\frac{k}{h-a}.....\left(ii\right)$

$\varphi +y=\pi \Rightarrow y=\pi -\varphi$

Given $m\tan x=\tan y$

$m\tan x=\tan ym\tan x=\tan (\pi -\varphi )m\tan x=-\tan \varphi$

substituting $\left(i\right)$ and $\left(ii\right)$

$m\frac{k}{h+a}=-\frac{k}{h-a}-hk-ak=mhk-mak-h-a=hm-amh+hm=am-a(1+m)h=(m-1)a$

Replacing $h$ by $x$

$(1+m)x=(m-1)a$

is the required locus of $A$