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
 

Given hyperbola $\frac{{\mathrm{x}}^2}{4{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{9{\mathrm{b}}^2}=1.$

Now equation of tangent at $\mathrm{P}\left(\theta \right)$ is

$$\begin{gathered} \Rightarrow \frac{\mathrm{x}}{2\mathrm{a}}\mathrm{sec\theta }-\frac{\mathrm{y}}{3\mathrm{b}}\mathrm{tan\theta }=1 \\ \Rightarrow \frac{\mathrm{x}}{{\displaystyle \frac{2\mathrm{a}}{\mathrm{sec\theta }}}}+\frac{\mathrm{y}}{{\displaystyle \frac{-3\mathrm{b}}{\mathrm{tan\theta }}}}=1 \end{gathered}$$

Given that $\frac{2\mathrm{a}}{\mathrm{sec\theta }}=1 \mathrm{and} \frac{-3\mathrm{b}}{\mathrm{tan\theta }}=1$

$$\begin{gathered} \Rightarrow \mathrm{sec\theta }=2\mathrm{a}, \mathrm{tan\theta }=-3\mathrm{b} \\ \mathrm{we} \mathrm{know} \mathrm{that} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow 4{\mathrm{a}}^2-9{\mathrm{b}}^2=1 \\ \therefore (\mathrm{a}, \mathrm{b}) \mathrm{lies} \mathrm{on} 4{\mathrm{x}}^2-9{\mathrm{y}}^2=1. \end{gathered}$$