Question

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

Answer 1

From the second relation, we get
$\frac{{\sin }^2\theta \sin \theta }{{\cos }^2\theta \cos \theta }=\frac{by}{ax}$
$\therefore \sin \theta =\frac{{\left(by\right)}^{1/3}}{{[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]}^{1/2}}$
$\cos \theta =\frac{{\left(ax\right)}^{1/3}}{{[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]}^{1/2}}$
Putting for $\sin \theta$ and $\cos \theta$
$\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=a^2-b^2$, we get
${[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]}^{1/2}[ax/{\left(ax\right)}^{1/3}+by/{\left(by\right)}^{1/3}]=a^2-b^2$
or ${[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]}^{1/2}[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]=a^2-b^2$
or ${[{\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}]}^{3/2}=a^2-b^2$
or ${\left(ax\right)}^{2/3}+{\left(by\right)}^{2/3}={(a^2-b^2)}^{2/3}$