# 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.

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.

• @ccorn yes thanks for your help! – David Quinn Jul 13 '16 at 6:11

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.