<s>[INST] There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$ Find the other side. There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$. Find the equation of 
other side.
I first let the sides are $OM$:  $y = m_1x$ and $ON:$ $y = m_2x$. Where $m_1m_2= \frac{-2h}{b}$ and $m_1 + 
m_2 = \frac{a}{b}$ 
Then I found the foot of the perpendicular of $(p, q)$ on $OM$. It was 
$$Q\left(\frac{p-m_1q}{m_1^2 + 1},m_1\frac{p-m_1q}{m_1^2 + 1}\right) $$ 
Now $Q$ is the midpoint of
$O$ and $M$. Thus I got $M$ is
$$\left(2\frac{p-m_1q}{m_1^2 + 1},2m_1\frac{p-m_1q}{m_1^2 + 1} \right)$$ 
Similarly, N is 
$$\left(2\frac{p-m_2q}{m_2^2 + 1} ,2m_2\frac{p-m_2q}{m_2^2 + 1} \right)$$
Using two point formula and making some manipulation, the equation of line $MN $ 
 i.e. the third side became 
$$\frac{(m_1^2 + 1)y- 2m_1p+2m_1^2q}{(m_1^2 + 1)x- 2p+2m_1q} =\frac{m_1m_2p}{(m_1+m_2)p+q}$$ 
From now where to go ?
 [/INST] Put $A(0,0),B(x,0),C(?,?)$. The coordinates of the vertex $C$ are intersection of the perpendiculars to the altitudes passing by $A$ and $B$, so we have after easy calculation $C(p,\frac{p(x-p)}{q})$.
Now one has $$(AC)^2=p^2+\left(\frac{p(x-p)}{q}\right)^2\\(BC)^2=(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2$$ The quotient $\dfrac{AC}{BC}$ is root of the given condition 
$ax^2+2hxy+by^2=0\iff a(\frac xy)^2+2h(\frac xy)+b=0$.
Consequently one has, putting $R=\dfrac{-2h\pm\sqrt{h^2-4ab}}{2a}$ (as data of problem)
$$\frac{(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2}{p^2+\left(\frac{p(x-p)}{q}\right)^2}=R^2$$
This quadratic equation gives the value of $x$ from which we have calculated the three vertices of the triangle determining this way the three sides, $x$ itself being the required side.
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