Question

Find the locus of the point of intersection of tangents to the circle $x=acos\theta ,$ $y=asin\theta$ at the point whose parametric angles differ by (i)$\frac{\pi }{3}$ (ii)$\frac{\pi }{2}.$

Answer 1

We have,

$x=a\cos \theta$

$y=a\sin \theta$

On differentiating and we get,

$\frac{dx}{d\theta }=-a\sin \theta ,\frac{dy}{d\theta }=a\cos \theta$

On

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\cos \theta }{-a\sin \theta }=-\cot \theta$

$\frac{dy}{dx}=-\cot \theta$

Equation of tangents

$y-y_1=\frac{dy}{dx}(x-x_1)$

$\Rightarrow y-a\sin \theta =-\frac{\cos \theta }{\sin \theta }(x-a\cos \theta )$

$\Rightarrow y\sin \theta -a{\sin }^2\theta =-x\cos \theta +a{\cos }^2\theta$

$\Rightarrow x\cos \theta +y\sin \theta =a({\sin }^2\theta +{\cos }^2\theta )$

$\Rightarrow x\cos \theta +y\sin \theta =a$

At point (1):-

$\frac{\pi }{3}$

$x\cos \theta +y\sin \theta =a$

$\Rightarrow x\cos \frac{\pi }{3}+y\sin \frac{\pi }{3}=a$

$\Rightarrow x\times \frac{1}{2}+y\times \frac{\sqrt{3}}{2}=a$

$\Rightarrow x+\sqrt{3}y=2a$

At point $\frac{\pi }{2}$

$x\cos \theta +y\sin \theta =a$

$\Rightarrow x\cos \frac{\pi }{2}+y\sin \frac{\pi }{2}=a$

$\Rightarrow x\times 0+y\times 1=a$

$\Rightarrow y=a$

Hence, this is the answer.