(x,y)∈R x^2 −xy−12y^2 =0 ((2x^2 +4y^2 )/(xy))=?


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Question Number 171289 by mathlove last updated on 11/Jun/22

$(x, y) \in R x^{2} - x y - 12 y^{2} = 0$ $\frac{2 x^{2} + 4 y^{2}}{x y} = ?$
	Answered by mr W last updated on 11/Jun/22

	$(x - 4 y) (x + 3 y) = 0$ $\Rightarrow x = 4 y o r x = - 3 y$ $\frac{2 x^{2} + 4 y^{2}}{x y} = \frac{2 {(4 y)}^{2} + 4 y^{2}}{(4 y) y} = 9 o r$ $\frac{2 x^{2} + 4 y^{2}}{x y} = \frac{2 {(- 3 y)}^{2} + 4 y^{2}}{(- 3 y) y} = - \frac{22}{3}$
	Commented by mathlove last updated on 11/Jun/22

	$t h a n k s$
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