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?
 A: $$(a+b)^2=4h^2 \Rightarrow a+b=\pm 2h$$
Let $$ax^2+2hxy+by^2=0$$
be factorized as $$(y-m_1x)(y-m_2x)=0$$
Then,
$$m_1+m_2=-\frac{2h}{a}$$
$$m_1m_2=\frac{b}{a}$$
We have,
$$1+\frac{b}{a}=\pm \frac{2h}{a}$$
So,
$$1+m_1m_2=\pm (m_1+m_2)$$
If either of $m_1$ or $m_2$ is zero, the other must be $\pm 1$, which means we are done.
If none of them are zero, let $m_1=km_2$.
$$1+km_2^2=\pm m_2(1+k)$$
$$km_2^2 \pm m_2(1+k) +1=0$$
which gives $m_2=\pm 1, \pm \frac{1}{k}$
which means $m_1=\pm \frac{1}{k},\pm 1$ correspondingly.
Thus at least one of them must bisect the coordinate axes.
A: 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.
A: I think this one is easier to work from the other direction.  If $(a+b)^2 = 4 h^2$, then $2 h = \pm (a + b)$.  Thus, the equation for the lines becomes
$$
a x^2 \pm (a + b) xy + b y^2 = 0 \quad \Rightarrow \quad (ax \pm by)(x \pm y) = 0.
$$
Since $(x \pm y)$ is one of the factors, the line $y = \pm x$ will be a solution to the equation.
A: 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 $-1$. 
A: Well... It seems it is asked to demonstrate that it works if $(a+b)^2=4h^2$, not if and only if...
So if you replace $h$ with $\pm \dfrac{a+b}{2}$ in your first equation you get the following:
$(x\pm y)(ax\pm by)=0$, which is your answer...
A: If one of the line of 
$$ax^2 +2hxy+by^2=0 \tag{1}$$
bisects the coordinate axes then the eqn. of that line might be $y=x$ OR, $y =-x$.
Combining these eqns we get $y=\pm x$ and substituting this value of $y$ in eqn. (1) we get
\begin{eqnarray}
 ax^2+2hx(\pm x)+by^2=0 \\
 ax^2±2hx^2+by^2=0 \\
 x^2(a±2h+b)=0 \\
 a±2h+b=0 \\
 a+b=±2h
\end{eqnarray}
Also, $(a+b)^2=4h^2$
