Question

The equation of locus of the point of intersection of tangents to the circle $x^2+y^2=1$ at the point whose parametric angles differ by ${60}^0$ is

• A. ${3x}^2+3y^2=1$
• B. $x^2+y^2=3$
• C. ${3x}^2+3y^2=4$
• D. None of these

Answer 1

Answer: (C)

${3x}^2+3y^2=4$

The point of intersection of the tangents at the points $P\left({\theta }_1\right)$ and $Q\left({\theta }_2\right)$ on the circle $x^2+y^2=1$ is given by

$x=\frac{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}=\frac{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{60}^0}{2}\right)}$

$y=\frac{a\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}=\frac{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\cos \left(\frac{{60}^0}{2}\right)}$

$\Rightarrow {\left(x\cos {30}^0\right)}^2+{\left(y\cos {30}^0\right)}^2=1$

$\Rightarrow (x^2+y^2)\frac{3}{4}=1\Rightarrow x^2+y^2=\frac{4}{3}$

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