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x-y-R-x-2-xy-12y-2-0-2x-2-4y-2-xy-




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

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