Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.

Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows;
here.
$$ax^2+2hxy+by^2=0$$
Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides :
$$ax+hy=\pm y\sqrt{h^2-ab}.$$
What should I do next?

If $ax^2+2hxy+by^2=0$ bisects the coordinate axes then the points $(x,x)$ and $(x,-x)$ belong to these lines. If we consider $(x,x)$ we get $$ax^2+2hx^2+bx^2=0$$ and if we consider $(x,-x)$ we get $$ax^2-2hx^2+bx^2=0.$$ If $x\ne 0$ then it is $$a+b=\pm 2h\implies (a+b)^2=4h^2.$$
Conversely, if $4h^2=(a+b)^2$ then, if $2h=a+b$ we have $ax^2+2hxy+by^2=(x+y)(ax+by)$ and if $2h=-(a+b)$ we have $ax^2+2hxy+by^2=(x-y)(ax+by).$  In any case, $ax^2+2hxy+by^2=0$ contains one of the lines $x+y=0$ or $x-y=0,$ which bisect the coordinate axis.
