Equation of Angle Bisectors

Combined equation of the two angle bisectors of ax² + 2hxy + by² = 0. The bisectors are always perpendicular to each other (their combined equation satisfies the perpendicularity condition: sum of x² and y² coefficients = 1 + (−1) = 0).

Let the two lines of $ax^2+2hxy+by^2=0$ be $l_1x+m_1y=0$ and $l_2x+m_2y=0$.

A point $P(x,y)$ lies on the angle bisectors iff it is equidistant from both lines:

$$\frac{l_1x+m_1y}{\sqrt{l_1^2+m_1^2}} = \pm\frac{l_2x+m_2y}{\sqrt{l_2^2+m_2^2}}$$

The combined equation of the two bisectors is obtained by multiplying these:

$$\frac{(l_1x+m_1y)^2}{l_1^2+m_1^2} = \frac{(l_2x+m_2y)^2}{l_2^2+m_2^2}$$

Working through the algebra using $l_1l_2 = a$, $l_1m_2+l_2m_1 = 2h$, $m_1m_2 = b$, $l_1^2+m_1^2+l_2^2+m_2^2 = (a+b) + \text{others}$... the standard result emerges:

$$\frac{x^2-y^2}{a-b} = \frac{xy}{h}$$

Key properties of the bisectors:

1. The two bisectors are always perpendicular to each other. Proof: the bisector equation has $x^2$ coefficient $= 1/(a-b)$ and $y^2$ coefficient $= -1/(a-b)$; their sum is zero — perpendicularity condition satisfied.
2. The bisectors of $ax^2+2hxy+by^2=0$ are the same as the bisectors of any pair of lines with the same angle — they depend only on $a$, $h$, $b$.
3. When $a = b$: the bisectors are $xy = 0$, i.e. the coordinate axes.
4. When $h = 0$: the bisectors are $x^2-y^2 = 0$, i.e. $y = \pm x$.