$\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=a^2-b^{2}and\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }=0then{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}=$

$\begin{array}{l}Givenequationsareax\sin \theta +by\cos \theta -(a^2-b^2)\sin \theta \cos \theta =0\\ andax{\sin }^3\theta -by{\cos }^3\theta =0,bysolvingweget\\ ax=\frac{(a^2-b^2){\cos }^3\theta \cos \theta \sin \theta }{{\cos }^3\theta \sin \theta +{\sin }^3\theta \cos \theta }=(a^2-b^2){\cos }^3\theta andbx=(a^2-b^2){\sin }^3\theta \\ \therefore {\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}={(a^2-b^2)}^{2/3}({\cos }^2\theta +{\sin }^2\theta )={(a^2-b^2)}^{2/3}.\end{array}$