If the product of the slopes of the tangents drawn from an external point P to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ is a constant ${\mathrm{k}}^2,$ then the locus of P is

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{let} \mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2} \mathrm{be} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{drawn} \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{then} \\ {({\mathrm{y}}_1-{\mathrm{mx}}_1)}^2={\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2 \\ \Rightarrow {{\mathrm{y}}_1}^2+{\mathrm{m}}^2{{\mathrm{x}}_1}^2-2{\mathrm{mx}}_1{\mathrm{y}}_1={\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2 \\ \Rightarrow ({{\mathrm{x}}_1}^2-{\mathrm{a}}^2){\mathrm{m}}^2-2{\mathrm{mx}}_1{\mathrm{y}}_1+{{\mathrm{y}}_1}^2+{\mathrm{b}}^2=0 \\ \mathrm{Now} {\mathrm{m}}_1{\mathrm{m}}_2=\frac{{{\mathrm{y}}_1}^2+{\mathrm{b}}^2}{{{\mathrm{x}}_1}^2-{\mathrm{a}}^2} \\ \Rightarrow {\mathrm{k}}^2=\frac{{{\mathrm{y}}_1}^2+{\mathrm{b}}^2}{{{\mathrm{x}}_1}^2-{\mathrm{a}}^2} \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{is} {\mathrm{y}}^2+{\mathrm{b}}^2={\mathrm{k}}^2({\mathrm{x}}^2-{\mathrm{a}}^2) \end{gathered}$$