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 difference of the base angles is given $(=\alpha )$.

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

Given $y-x=\alpha$

$\Rightarrow y=x+\alpha \varphi +y=\pi \varphi =\pi -y\varphi =\pi -(x+\alpha )\tan \varphi =\tan \left[\pi -(x+\alpha )\right]\tan \varphi =-\tan (x+\alpha )\tan \varphi =-\tan (\alpha +x)\tan \varphi =-\frac{\tan \alpha +\tan x}{1-\tan \alpha \tan x}$

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

$\frac{k}{h-a}=-\frac{\tan \alpha +\frac{k}{h+a}}{1-\tan \alpha \frac{k}{h+a}}$

$\frac{k}{h-a}=-\frac{h\tan \alpha +a\tan \alpha +k}{h+a-k\tan \alpha }$

$hk+ak-k^2\tan \alpha =-h^2\tan \alpha -ah\tan \alpha -hk+ah\tan \alpha +a^2\tan \alpha +ak$

$-h^2\tan \alpha +k^2\tan \alpha -2hk=-a^2\tan \alpha$

$h^2-k^2+2hk\cot \alpha =a^2$

$x^2+2xy\cot \alpha -y^2=a^2$

is the required locus of vertex $A$