Question Number 171289 by mathlove last updated on 11/Jun/22
$$\left({x},{y}\right)\in{R}\:\:\:{x}^{\mathrm{2}} −{xy}−\mathrm{12}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }{{xy}}=? \\ $$
Answered by mr W last updated on 11/Jun/22
$$\left({x}−\mathrm{4}{y}\right)\left({x}+\mathrm{3}{y}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{4}{y}\:{or}\:{x}=−\mathrm{3}{y} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }{{xy}}=\frac{\mathrm{2}\left(\mathrm{4}{y}\right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }{\left(\mathrm{4}{y}\right){y}}=\mathrm{9}\:{or} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }{{xy}}=\frac{\mathrm{2}\left(−\mathrm{3}{y}\right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }{\left(−\mathrm{3}{y}\right){y}}=−\frac{\mathrm{22}}{\mathrm{3}} \\ $$
Commented by mathlove last updated on 11/Jun/22
$${thanks} \\ $$