A tangent to the curve $9{\mathrm{b}}^2{\mathrm{x}}^2 – 4{\mathrm{a}}^2{\mathrm{y}}^2 = 36{\mathrm{a}}^2{\mathrm{b}}^2$ makes intercepts of unit length on each of the coordinate axes, then the point (a, b) lies on

$$\begin{gathered} Given hyperbola 9b^2{x}^2-4a^2{y}^2=36a^2{b}^2 \\ \Rightarrow \frac{{\mathrm{x}}^2}{4{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{9{\mathrm{b}}^2}=1 \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P}\left(\mathrm{\theta }\right) \mathrm{is} \\ \frac{\mathrm{x}}{2\mathrm{a}} \mathrm{Sec\theta }-\frac{\mathrm{y}}{3\mathrm{b}}\mathrm{tan\theta }=1 \\ \Rightarrow \frac{x}{2a \cos \theta }+\frac{y}{-3b cot\theta }=1 \\ \mathrm{Given} \mathrm{that} 2a \cos \theta =1 \mathrm{and} -3b cot\theta =1 \\ \Rightarrow \mathrm{sec\theta }=2\mathrm{a}, \mathrm{tan\theta }=-3\mathrm{b} \\ \mathrm{Now} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow 4{\mathrm{a}}^2-9{\mathrm{b}}^2=1 \\ \therefore (\mathrm{a},\mathrm{b}) \mathrm{lies} \mathrm{an} 4{\mathrm{x}}^2-9{\mathrm{y}}^2=1 \end{gathered}$$