Question

Find the locus of the point of intersection of tangents to the circle $x=a\cos \theta ,y=a\sin \theta$ at the point whose parametric angles differ (i) by $\pi /3$, (ii) by $\pi /2$.

Answer 1

Let the parametric angles be $\alpha$ and $\alpha +\pi /3$ and equation of the circle is $x^2+y^2=a^2$.
Tangent at $\alpha$, i.e. $(a\cos \alpha ,a\sin \alpha )$ is
$x\cos \alpha +y\sin \alpha =a.....\left(1\right)$
Tangent at $\alpha +\pi /3$ is
$x\cos (\alpha +\pi /3)+y\sin (\alpha +\pi /3)=a$.
or $x(\cos \alpha \cos \pi /3-\sin \alpha \sin \pi /3)+y(\sin \alpha \cos \pi /3+\cos \alpha \sin \pi /3)=a$.
or $\frac{1}{2}(x\cos \alpha +y\sin \alpha )-\left(\sqrt{3}/2\right)(x\sin \alpha -y\cos \alpha )=a$
or $\frac{1}{2}a-\frac{\sqrt{3}}{2}(x\sin \alpha -y\cos \alpha )=a$ by $\left(1\right)$
or $x\sin \alpha -y\cos \alpha =-a/\sqrt{3}.....\left(2\right)$
In order to find the locus of the point of intersection we have to eliminate the parameter $\alpha$ for which we square and add $\left(1\right)$ and $\left(2\right)$.
$\therefore x^2({\cos }^2\alpha +{\sin }^2\alpha )+y^2({\sin }^2\alpha +{\cos }^2\alpha )$
$=a^2+a^2/3=4a^2/3$
or $3x^2+3y^2=4a^2$
When the parametric angles differ by $\pi /2$, then the $2nd$ tangent at $(\alpha +\frac{\pi }{2})$ is
$-x\sin \alpha +y\cos \alpha =a....\left(3\right)$
The locus is found by squaring and adding $\left(1\right)$ and $\left(3\right)$ as $x^2+y^2=2a^2$.