# Condition for the pair of lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ to be parallel

by ss1729   Last Updated August 13, 2019 20:20 PM - source

Show that the condition for the pair of lines $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ to be parallel is $$ab=h^2$$ and $$bg^2=af^2$$ or $$\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$$.

$$ab=h^2$$ condition is understandable as the acute angle between the lines $$\tan\theta=\dfrac{2\sqrt{h^2-ab}}{a+b}$$ can be obtained from the corresponding lines going through the origin $$ax^2+2hxy+by^2=0$$.

In a similar post Deriving conditions for a pair of straight lines to be parallel, it is i think attempted to prove by taking partial derivatives with respect to $$x$$ and $$y$$, and taking $$𝑎𝑥+ℎ𝑦+𝑔=0$$ and $$ℎ𝑥+𝑏𝑦+𝑓=0$$ to be coincident.

I simply do not understand the logic behind such an attempt ?

And how do I prove that the pair of lines represented by the second order equation are parallel and not coincident ?

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Suppose the given equation factors as $$fg=0$$ Partially differentiating by $$x$$ gives $$f_xg + fg_x=0$$ Partially differentiating by $$y$$ gives $$f_yg+fg_y = 0$$ Just notice that any point which lies on both $$f$$ and $$g$$ also satisfies both above equations.