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?
$$(a+b)^2=4h^2 \Rightarrow a+b=\pm 2h$$
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.
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, $(a+b)^2=4h^2$