Two tangents to the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$having slopes m₁ and m₂ intersect the axes at four concyclic points. Then the value of m₁ m₂ is

$$\begin{gathered} Given hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ Let m_1,m_2 be slopes of two \tan gents \end{gathered}$$

Now Equations of tangents are $\mathrm{y}={\mathrm{m}}_1\mathrm{x}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}_1-{\mathrm{b}}^2}$;

$\mathrm{y}={\mathrm{m}}_2\mathrm{x}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}_2^2-{\mathrm{b}}^2}$

Since four points are concyclic so four points lie on circle 

sub x=0 and y=0  then we four points

$$\begin{gathered} A=(\frac{-\sqrt{m_1{a}^2-b^2}}{m_1},0),B=(0,\sqrt{a^2{m_1}^2-b^2}) and \\ C=(\frac{-\sqrt{m_2{a}^2-b^2}}{m_2},0),D=(0,\sqrt{a^2{m_2}^2-b^2}) \\ Since four points are concyclic then \\ OA\times OC=OB\times OD \\ \Rightarrow \frac{-\sqrt{m_1{a}^2-b^2}}{m_1}\times \frac{-\sqrt{m_2{a}^2-b^2}}{m_2}=\sqrt{a^2{m_1}^2-b^2}\times \sqrt{a^2{m_2}^2-b^2} \\ \Rightarrow \frac{1}{m_1{m}_2}=1 \\ \Rightarrow m_1{m}_2=1 \end{gathered}$$