set of lines through origin that bisects both lines of $x^2 - pxy + y^2$

Question

I've been struggling with this for quite sometime and looks like the answer is $x^2 +1/p *xy +y^2$. Not sure how they got this.

Answer 1

Assuming that you’re looking for the angle bisectors, we can view this pair of lines as a degenerate hyperbola: the angle bisectors are then its axes. The matrix of $x^2-pxy+y^2$ is $$Q=\begin{bmatrix} 1&-\frac p2\\-\frac p2&1 \end{bmatrix}$$ and the principal axes are its eigenvectors, which we can find by inspection: $$Q[1,1]^T=\left[1-\frac p2,1-\frac p2\right]^T = \left(1-\frac p2\right)[1,1]^T \\ Q[1,-1]^T=\left[1+\frac p2,-1-\frac p2\right]^T = \left(1+\frac p2\right)[1,-1]^T,$$ i.e., the angle bisectors are $x+y=0$ and $x-y=0$.

Answer 2

$x^2 -p xy +y^2$. is your initial equation.

Assume 2 lines which intersect origin be of form x+ay = 0

=>x = -ay

Now since this lines intersect above set, it should satisfy that euqation.

Substituting value of x, we have

$(-ay)^2 -p(-ay)y +y^2 = 0$

$a^2y^2 +pay^2 +y^2 = 0$

$y^2(a^2 +pa +1) = 0$

Either y =0 or the equation in bracket. Discarding trivial solution we see that our equation is quadratic in $a$ and roots are

a1=$\frac{-p+\sqrt{p^2-4}}{2}$ and a2=$\frac{-p-\sqrt{p^2-4}}{2}$

Hence with 2 roots we have 2 equations which pass through origin and intersect the given pair.

Multiply these 2 equations

(x+a1y)*(x+a2y)=0

Solve it and you will get the desired result

QED

Answer 3

Better to write as

$$ x^2-pxy+y^2= a^2,\, y=m\,x $$

where it is cut by a straight line with slope $p$ through origin/center.

Plug in and solve for $x$:

$$ x_{1,2}= \pm \dfrac{a}{\sqrt{1-p\,m +m^2}}$$

which are equal in magnitude but opposite in sign as they lie on either side of $y-$ axis. This happens for $y_{1,2}$ also, when the roots lie on either side of $y-$ axis. So he enclosed segment of the cutting straight line is bisected at the origin.