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My question is:

How to solve this equation:


with respect to $x$ and $y$ in the same time. Here $a,b,c$ are real constants.

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You mean you want, say, $y$ as a fucntion of $x$? – Ron Gordon Jan 4 '13 at 15:36
Yes. This is the case along solving it with respect to $(x,y)$ – ZE1 Jan 4 '13 at 15:38
up vote 2 down vote accepted

$$ax²+by²+cxy=0/y^2,y\neq0$$ $$a(x/y)^2+c(x/y)+b=0,a\neq0$$

$$x/y=\frac{-c+\sqrt{c^2-4ab}}{2a},x/y=\frac{-c-\sqrt{c^2-4ab}}{2a}$$ if $y=0$ then $ax^2=0$ follow that $x=0$

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If $a\neq 0$, the equation $$ax^2+(cy)x+by^2=0$$ is in effect a quadratic with respect to $x$. The discriminant for example is $\Delta=(cy)^2-4aby^2$. This will allow you to express $x$ in terms of $y$ only (if the discriminant is non-negative). Alternatively you can consider the equation $$by^2+(cx)y+ax^2=0$$ as a quadratic with respect to $y$ (if $b\neq 0$) and solve for $y$. The cases $a=0$ or $b=0$ are very simple

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Yes and thank you. – ZE1 Jan 4 '13 at 15:43

use the quadratic formula to find $y$:

$$y = \frac{-c x \pm \sqrt{c^2 - 4 a b } |x|}{2 b} $$

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You may want to clarify that this is true if $c^2-4ab\ge 0$ – Nameless Jan 4 '13 at 15:46

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