Question

If $\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=a^2-b^2$ ,and $\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }=0$,

prove that $(ax{)}^{2/3}+(by{)}^{2/3}=(a^2-b^2{)}^{2/3}$

Answer 1

We have: $\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }=0$

$\Rightarrow ax{\sin }^3\theta -by{\cos }^3\theta =0$

$\Rightarrow \frac{{\sin }^3\theta }{by}=\frac{{\cos }^3\theta }{ax}$

$\Rightarrow {\left(\frac{{\sin }^3\theta }{by}\right)}^{2/3}={\left(\frac{{\cos }^3\theta }{ax}\right)}^{2/3}$

$\Rightarrow \frac{{\sin }^2\theta }{(by{)}^{2/3}}=\frac{{\cos }^2\theta }{(ax{)}^{2/3}}$

$\Rightarrow \frac{{\sin }^2\theta }{(by{)}^{2/3}}-\frac{{\cos }^2\theta }{(ax{)}^{2/3}}-\frac{{\sin }^2\theta +{\cos }^2\theta }{(by{)}^{2/3}+(ax{)}^{2/3}}$

$\Rightarrow \frac{{\sin }^2\theta }{(by{)}^{2/3}}=\frac{{\cos }^2\theta }{(ax{)}^{2/3}}=\frac{1}{(ax{)}^{2/3}+(by{)}^{2/3}}$

$\Rightarrow {\sin }^2\theta =\frac{(by{)}^{2/3}}{(ax{)}^{2/3}+(by{)}^{2/3}}$

${\cos }^2\theta =\frac{(ax{)}^{2/3}}{(ax{)}^{2/3}+(by{)}^{2/3}}$

$\Rightarrow \sin \theta =\frac{(by{)}^{1/3}}{\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}}$ and $\cos \theta =\frac{(ax{)}^{1/3}}{\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}}$

Substituting these value in

$\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=a^2-b^2$, we get

$(ax{)}^{2/3}\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}+(by{)}^{2/3}\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}=a^2-b^2$

$\Rightarrow \left[\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}\right]\left[\right(ax{)}^{2/3}+\left(by{)}^{2/3}\right]=a^2-b^2$

$\Rightarrow {\left[\right(ax{)}^{2/3}+\left(by{)}^{2/3}\right]}^{3/2}=a^2-b^2$

$\Rightarrow (ax{)}^{2/3}+(by{)}^{2/3}=(a^2-b^2{)}^{2/3}$