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 product of the tangents of the base angles is given $(=\lambda )$.

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 $\tan x\tan y=\lambda$

$\tan x\tan (\pi -\varphi )=\lambda \tan x(-\tan \varphi )=\lambda -\tan x\tan \varphi =\lambda$

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

$-\frac{k}{h+a}\times \frac{k}{h-a}=\lambda \frac{k^2}{h^2-a^2}=-\lambda k^2=-\lambda h^2+a^2\lambda k^2+\lambda h^2=a^2\lambda$

Replacing $h$ by $x$ and $y$ by $k$

$y^2+\lambda x^2=a^2\lambda$

is the required locus of vertex $A$