If the tangent at any point on the curve ${\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{2/3}+{\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}^{2/3}=1$ makes the intercepts $\mathrm{p}, \mathrm{q}$ on the axes then $\frac{{\mathrm{p}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{q}}^2}{{\mathrm{b}}^2}=$

$$\begin{gathered} \mathrm{x}=\mathrm{a} {\cos }^3\mathrm{\theta } \mathrm{y}=\mathrm{b} {\sin }^3\mathrm{\theta } \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3\mathrm{b} {\sin }^2\mathrm{\theta } \mathrm{cos\theta }}{-3\mathrm{a} {\cos }^2\mathrm{\theta } \mathrm{sin\theta }}=\frac{-\mathrm{b} \mathrm{sin\theta }}{\mathrm{a} \mathrm{cos\theta }} \end{gathered}$$

eq of Tangent is

$\begin{array}{rcl}(\mathrm{b} \mathrm{sin\theta })\mathrm{x}+(\mathrm{a} \mathrm{cos\theta }) \mathrm{y}& =& \mathrm{ab} \mathrm{sin\theta } \mathrm{cos\theta }\\ \frac{\mathrm{x}}{\mathrm{a} \mathrm{cos\theta }}+\frac{\mathrm{y}}{\mathrm{b} \mathrm{sin\theta }}& =& 1\\ \mathrm{P}& =& \mathrm{a} \mathrm{cos\theta } \mathrm{q}=\mathrm{b} \mathrm{sin\theta }\\ {\left(\frac{\mathrm{p}}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{q}}{\mathrm{b}}\right)}^2& =& 1\end{array}$