Question

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

Answer 1

Let one of the points on the circle be $A(a\cos \theta ,a\sin \theta )$

Then the other point will be $B\left(a\cos \right(\theta +\pi ),a\sin (\theta +\pi \left)\right)$

Therefore equation of tangent at $A=x\cos \theta +y\sin \theta =a$...(1)

Equation of tangent at $B=x\cos (\theta +\pi )+y\sin (\theta +\pi )=a$...(2)

From (2)

$\Rightarrow x[\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta ]+y[\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta ]=a$

$A(a\cos \theta ,a\sin \theta )$

$B\left(a\cos \right(\theta +\pi ),a\sin (\theta +\pi \left)\right)$

$A=x\cos \theta +y\sin \theta =a$...(1)

$\Rightarrow x[\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta ]+y[\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta ]=a$

$\Rightarrow \frac{1}{2}(x\cos \theta +y\sin \theta )-\frac{\sqrt{3}}{2}(x\sin \theta -y\cos \theta )=a$

$\Rightarrow (x\sin \theta -y\cos \theta )=\frac{-a}{\sqrt{3}}$...(3)

Square and add (1) and (3)

$(x\cos \theta +y\sin \theta {)}^2+(x\sin \theta -y\cos \theta {)}^2=a^2+\frac{a^2}{3}$

$\Rightarrow 3x^2+3y^2=4a^2$

The locus of the point of intersection of the tangents $3x^2+3y^2-4a^2=0$