The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{5}=1$ meets x-axis and y-axis at A and B respectively. Then ${\left(\mathrm{OA}\right)}^2-{\left(\mathrm{OB}\right)}^2,$where O is the origin, equals

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{5}=1 \\ where {\mathrm{a}}^2=4, {\mathrm{b}}^2=5 \\ \mathrm{Now} \mathrm{e}=\sqrt{\frac{4+5}{4}}=\frac{3}{2} \\ \mathrm{One} \mathrm{end} \mathrm{of} \mathrm{latus} \mathrm{rectum} \left(\mathrm{P}\right)=\left(\mathrm{ae}, \frac{{\mathrm{b}}^2}{\mathrm{a}}\right) \\ =\left(3, \frac{5}{2}\right) \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P} \mathrm{is} \\ \frac{\mathrm{x}\left(3\right)}{4}-\frac{\mathrm{y}\left({\displaystyle \frac{5}{2}}\right)}{5}=1 \\ \Rightarrow \frac{3\mathrm{x}}{4}-\frac{\mathrm{y}}{2}=1 \\ \Rightarrow \frac{\mathrm{x}}{{\displaystyle \frac{4}{3}}}+\frac{\mathrm{y}}{-2}=1 \\ \mathrm{x}-\mathrm{intercert} \left(\mathrm{OA}\right)=\frac{4}{3} \\ y-\mathrm{intercert} \left(\mathrm{OB}\right)=-2 \\ \mathrm{Now} {\mathrm{OA}}^2-{\mathrm{OB}}^2=\frac{16}{9}-4=\frac{-20}{9} \end{gathered}$$