$$\frac{x}{y}=\frac{5}{2}; 3x^{2}+5y^{2}; 3x^{2}-5y^{2}$$
$x=\frac{5y}{2}\text{, }y=-\frac{2\sqrt{95z}}{95}\text{, }z>0$
$x=\frac{5y}{2}\text{, }y=\frac{2\sqrt{95z}}{95}\text{, }z>0$
