The locus of the point of intersection of two tangents to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ which make an angle ${60}^0$ with one another is

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{to} \mathrm{the} \mathrm{given} \mathrm{hyperbola} \mathrm{with} \mathrm{slope} \text{'}\mathrm{m}\text{'} \mathrm{is} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \\ \Rightarrow {(\mathrm{y}-\mathrm{mx})}^2 ={\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2 \\ \Rightarrow {\mathrm{y}}^2+{\mathrm{m}}^2{\mathrm{x}}^2-2\mathrm{mxy}-{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2=0 \\ \Rightarrow ({\mathrm{x}}^2-{\mathrm{a}}^2){\mathrm{m}}^2-2\mathrm{mxy}+{\mathrm{y}}^2+{\mathrm{b}}^2=0 \\ It is a quadratic equation in m.It has two roots \mathrm{let} it be {\mathrm{m}}_1, {\mathrm{m}}_2 \\ where m_1+m_2=\frac{2m}{x^2-a^2}, m_1.m_2=\frac{{\mathrm{y}}^2+{\mathrm{b}}^2}{{\mathrm{x}}^2-{\mathrm{a}}^2} \\ \mathrm{Now} \tan \mathrm{\theta }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right| \\ \Rightarrow \sqrt{3}=\pm \frac{\sqrt{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2-4{\mathrm{m}}_1{\mathrm{m}}_2}}{1+{\mathrm{m}}_1{\mathrm{m}}_2} \\ \Rightarrow 3=\frac{{\left({\displaystyle \frac{2\mathrm{xy}}{{\mathrm{x}}^2-{\mathrm{a}}^2}}\right)}^2-4\left({\displaystyle \frac{{\mathrm{y}}^2+{\mathrm{b}}^2}{{\mathrm{x}}^2-{\mathrm{a}}^2}}\right)}{{\left(1+\left({\displaystyle \frac{{\mathrm{y}}^2+{\mathrm{b}}^2}{{\mathrm{x}}^2-{\mathrm{a}}^2}}\right)\right)}^2} \\ \mathrm{on} \mathrm{simplifying} \mathrm{we} \mathrm{get} \\ 3{({\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{a}}^2+{\mathrm{b}}^2)}^2=4({\mathrm{a}}^2{\mathrm{y}}^2-{\mathrm{b}}^2{\mathrm{x}}^2+{\mathrm{a}}^2{\mathrm{b}}^2) \end{gathered}$$