<s>[INST] 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?
 [/INST] Let $y=mx$ be the required line. Then we can rewrite the given pair of lines as $$bm^2+2hm+a=0$$.  
$$\implies m=\frac{-2h\pm\sqrt{4h^2-4ab}}{2b}$$   
$$(a+b)^2=4h^2 \implies 4h^2-4ab=(a-b)^2$$ 
Substituting in the value of $m$, $$m=\frac{-\vert a+b\vert\pm\vert a-b\vert}{2b}$$
For all values of $a,b$, the above will always gives at least $1$ or $-1$. 
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