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Suppose you have a pair of lines passing through origin, $ax^2 + 2hxy +by^2 = 0$, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for $2$ separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.

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Let's consider an easier case, when the given lines have the form $y=mx$ and $y=nx$ (with $m\ne n$, of course).

The points on the angle bisectors satisfy the conditions that their distance from the two lines is the same. So you have, for such a point $(x,y)$, $$ \frac{|y-mx|}{\sqrt{1+m^2}}=\frac{|y-nx|}{\sqrt{1+n^2}} $$ By squaring, we get $$ (y-mx)^2(1+n^2)=(y-nx)^2(1+m^2) $$ and expanding $$ (y-mx)^2-(y-nx)^2=m^2(y-nx)^2-n^2(y-mx)^2. $$ The left hand side becomes $$ (y-mx+y-nx)(y-mx-y+nx)=-(m-n)(2y-(m+n)x)x $$ and the right hand side is $$ (my-mnx+ny-mnx)(my-mnx-ny+mnx)=(m-n)((m+n)y-2mnx)y $$ Since $m\ne n$, we can factor out $m-n$ on both sides, getting $$ -2xy+(m+n)x^2=(m+n)y^2-2mnxy $$ that can be written $$ (m+n)x^2-2(1-mn)xy-(m+n)y^2=0 $$ If you consider the equation $by^2+2hxy+ax^2=0$ in the unknown $y$, you can see that $m$ and $n$ are the roots of it (when $b\ne 0$). Thus $m+n=-2h/b$ and $mn=a/b$, so you can rewrite the above equation in the form $$ \frac{-2h}{b}x^2-2\left(1-\frac{a}{b}\right)xy-\frac{-2h}{b}y^2=0 $$ which is $$ hx^2-(a-b)xy-hy^2=0 $$ Note that it can't be both $h=0$ and $a=b\ne0$, because in this case the original equations would be $x^2+y^2=0$ (the isotropic lines, if you consider the complex plane).

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  • $\begingroup$ Sweet answer!!! $\endgroup$ – user34304 May 8 '14 at 3:04
  • $\begingroup$ hey, since this is a degenerate hyperbola, would this also give me the axis of a general hyperbola $\endgroup$ – Vrisk Oct 26 '17 at 9:04
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    $\begingroup$ @Vrisk You need to use the asymptotes. $\endgroup$ – egreg Oct 26 '17 at 9:11

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