Two fixed points are A(a, 0) and B(-a, 0). If $\angle A-\angle B=\theta$, then the locus of point C of triangle ABC will be

The correct option is D ($x^2-y^2+2xycot\theta =a^2$.)

Given $\angle A-\angle B=\theta \Rightarrow tan(A-B)=tan\theta$ .............(i)

In right angled triangle CDA, tan A = $\frac{k}{a-h}$

and similarly in triangle CDB, tan B = $\frac{k}{a+h}$

Also from (i), $\frac{tanA-tanB}{1+tanA.tanB}=tan\theta$

Substituting the values of tan A and tan B, we get

$\Rightarrow \frac{\frac{k}{a-h}-\frac{k}{a+h}}{1+\left(\frac{k}{a-h}\right)\left(\frac{k}{a+h}\right)}=\tan \theta$

$\Rightarrow \frac{k(a+h-a+h)}{a^2-h^2+k^2}=\tan \theta$

$\Rightarrow \frac{2hk}{a^2-h^2+k^2}=\frac{1}{\cot \theta }$

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

$\Rightarrow h^2-k^2+2hkcot\theta =a^2$

Hence the locus is $x^2-y^2+2xycot\theta =a^2$.