Question

Find the locus of the intersection of tangents if the sum of the eccentric angles of their points of contact be equal to a constant angle $2\alpha$.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Let $(h,k)$ be the point of intersection of tangents.

Equation of chord of contact is $T=0$

$\frac{hx}{a^2}+\frac{ky}{b^2}-1=\mathrm{0.....}\left(i\right)$

Equation of chord when eccentric angles of ends points is given is

$\frac{x}{a}\cos \left(\frac{\theta +\varphi }{2}\right)+\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$

$\frac{x}{a}\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}+\frac{y}{b}\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}=1$ ..... $\left(ii\right)$

Both $\left(i\right)$ and $\left(ii\right)$ are chord of contact w.r.t. to $(h,k)$, so by comparing both the equations.

$\frac{h}{a^2}=\frac{1}{a}\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}\Rightarrow h=a\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}......\left(iii\right)\frac{k}{b^2}=\frac{1}{b}\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}\Rightarrow k=b\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}........\left(iv\right)$

Dividing $\left(iv\right)$ by $\left(iii\right)$, we get
$\frac{k}{h}=\frac{b}{a}\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}\times \frac{\cos \left(\frac{\theta -\varphi }{2}\right)}{\cos \left(\frac{\theta +\varphi }{2}\right)}$

$\Rightarrow ak=bh\tan \left(\frac{\theta +\varphi }{2}\right)$

Given $\theta +\varphi =2\alpha$

$\Rightarrow ak=bh\tan \alpha ay=bx\tan \alpha$

is the required equation of locus.