Sum of slopes between points of concurrent normals to hyperbola is zero Let $ A_1, A_2, A_3, A_4 $ be four points on the hyperbola $xy = 1$. Suppose that the
normals to the hyperbola at these four points are concurrent, i.e. they intersect in a
single point. Prove that the sum of slopes of the lines $ A_i A_j  
, i < j$ is zero.
What I tried: 
Let $ A_i=(a_i,a_i^{-1}) $. Then the equation of the normal line through $A_i$ is $$N(A_i)=a_i^{2}(x-a_i)+a_i^{-1} .$$
I cannot seem to find an example of four points that satisfy the above condition without one point being (1,1). That is, $$A_1=(1,1) \; and \;  N(A_1)=x.$$ This case forces the other three points to be in very nice locations. Namely,
$$A_2=(-1,-1)$$ and if $A_3=(a_3,a_3^{-1}) $ is a point in the $1^{st}$ or $3^{rd}$ quadrant, then $A_4=(a_3^{-1},a_3)$. Of course, since the lines don't intersect in exactly one point, this doesn't satisfy my hypothesis.
My question is this: 
Given some point $p$, how does one construct normals that pass through $p$? I've read that at most four normals can be drawn through any given point, but can't seem to find a proof of this.
 A: Any point on the hyperbola can be written as $\left(t, \frac{1}{t}\right)$. The equation of the normal at $t$ is 
\begin{equation*}
y - \frac{1}{t} = t^2(x-t)
\end{equation*}
If this passes through a given point $(x_1,y_1)$, we have 
\begin{equation*}
y_1 - \frac{1}{t} = t^2(x_1-t)
\end{equation*}
which can be written as
\begin{equation*}
t^4 - t^3x_1 + ty_1 -1 = 0
\end{equation*}
This in general has four roots. The roots give  the coordinates of the feet of the normals from the given point.
A: The equation of the normal at the point $(t,\frac 1t)$ is 
$$y-\frac 1t=t^2(x-t)$$
So the (in general) four values of $t$ for which the normal passes through $(p,q)$ satisfy the quartic equation $$t^4-pt^3+qt-1=0$$
Since the coefficient of $t^2$ is zero, the sum of the products of the roots taken in pairs is also zero, i.e. $$\sum t_it_j=0$$
However, the gradient of the line joining a typical pair of points is $$\frac{\frac{1}{t_i}-\frac{1}{t_j}}{t_i-t_j}=-\frac{1}{t_it_j}$$
Multiplying the sum of these terms by the product of all four roots, which is $-1$, then gives the required result.
