Tangents to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ make angles ${\mathrm{\theta }}_1, {\mathrm{\theta }}_2$ with the transverse axis. If ${\mathrm{\theta }}_1, {\mathrm{\theta }}_2$ are complementary then the locus of the point of  intersection of the tangents is

let $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}$ be equation of tangents

let p$(x_1,y_1) be point of intersection of \tan gents$

$$\begin{gathered} \Rightarrow {({\mathrm{y}}_1-{\mathrm{mx}}_1)}^2={\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2 \\ \Rightarrow {\mathrm{m}}^2({{\mathrm{x}}_1}^2-{\mathrm{a}}^2)-2\mathrm{mxy}+{{\mathrm{y}}_1}^2+{\mathrm{b}}^2=0 \\ let m_{1,}{m}_2 are roots of above equation then \\ {m}_1{\mathrm{m}}_2=\frac{{{\mathrm{y}}_1}^2+{\mathrm{b}}^2}{{{\mathrm{x}}_1}^2-{\mathrm{a}}^2} \\ \Rightarrow \tan {\theta }_1 {\mathrm{tan\theta }}_2=\frac{{{\mathrm{y}}_1}^2+{\mathrm{b}}^2}{{{\mathrm{x}}_1}^2-{\mathrm{a}}^2} \\ \Rightarrow 1=\frac{{{\mathrm{y}}_1}^2+{\mathrm{b}}^2}{{{\mathrm{x}}_1}^2-{\mathrm{a}}^2}(\because {\mathrm{\theta }}_1\& {\mathrm{\theta }}_2 \mathrm{are} \mathrm{complementary} \mathrm{angles}) \\ \Rightarrow {{\mathrm{x}}_1}^2-{\mathrm{a}}^2={{\mathrm{y}}_1}^2+{\mathrm{b}}^2 \\ \Rightarrow {{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{p} \mathrm{is} {\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \end{gathered}$$