Q.If the pair of straight lines $x^2 - 2kxy - y^2 = 0$ bisects the angle between the pair of straight lines $x^2 - 2lxy - y^2 = 0$, show that the latter pair also bisects the angle between the former.

For a homogeneous pair $Ax^2+2Hxy+By^2=0$, the pair of angle bisectors is $\dfrac{x^2-y^2}{A-B}=\dfrac{xy}{H}$.

Step 1. Find the bisector of $x^2-2lxy-y^2=0$. Here $A=1,\ H=-l,\ B=-1$, so $A-B=2$:

Step 2. Impose the hypothesis: this bisector equals the pair $x^2-2kxy-y^2=0$. Two homogeneous quadratics represent the same pair of lines exactly when their coefficients are proportional:

Matching $x^2$: $l=t$. Matching $y^2$: $-l=-t$ (consistent, $t=l$). Matching $xy$: $2=t(-2k)=-2kl$ (using $t=l$), so

This is exactly the condition under which the pair $x^2-2kxy-y^2=0$ is the bisector of $x^2-2lxy-y^2=0$.

Step 3. Now find the bisector of $x^2-2kxy-y^2=0$ (the other direction). Here $A=1,\ H=-k,\ B=-1$, $A-B=2$:

$\frac{x^2-y^2}{2}=\frac{xy}{-k}\ \Longrightarrow\ kx^2+2xy-ky^2=0.$ …

• Treating the two homogeneous equations as literally equal instead of proportional (missing the scalar $t$), which can hide a sign error …