Find the single equation of the two lines through the origin and perpendicular to the each lines represented by $ax^2+2hxy+by^2=0$

I tried the factorization of the given equation but it was fail..


Given $ax^2 + 2hxy + by^2 = 0$ is equation of two lines passing through origin.
Let the equation of lines be $y = m_1x$ and $y = m_2x$

Dividing the equation by $x^2$
$$b(\frac yx)^2 + 2h(\frac yx) + a = 0$$ We know $\frac yx$ is the slope of the lines, say $m$
Thus, $$bm^2 + 2hm + a = 0$$
Its clearly a quadratic equation. Solving $$m = \frac{-2h \pm \sqrt{4h^2 - 4ab}}{2b}$$ or $$m = \frac{-h \pm \sqrt{h^2 - ab}}{b}$$ Thus equation being $$y = \frac{-h \pm \sqrt{h^2 - ab}}{b}x$$

  • $\begingroup$ for $y$ we must have $$y\ne 0$$ but what is in the other case? $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '16 at 17:04
  • $\begingroup$ I should've divided by $x$. Now if $x = 0$, then $x$ must be a factor of your equation i.e. $b=0$. You've $(ax+2hy)x = 0$. Thus either $x=0$ or $ax+2hy=0$. Put $b$ in the last equation, you'll divide by 0. Agreeing the fact that slope is not defined. $\endgroup$ – dark32 Jan 1 '16 at 17:19

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