I came across the Wikipedia page on conjugate diameters of ellipses, circles and hyperbolas, stating

[T]wo diameters of a conic section are said to be conjugate if each chord parallel to one diameter is bisected by the other diameter.

There is a statement that the conjugate of a diameter of a rectangular hyperbola is a reflection across an asymptote. The entry further links to this page that describes it as the hyperbolic orthogonality property of two lines. Visually, it is easy and intuitive to make this deduction. However, the pages do not include any proofs - perhaps it is too lengthy. Where can I find a proof of this property?

  • 2
    $\begingroup$ Seems pretty easy to prove analytically: Consider the hyperbola $xy=1$ and the diameter $y=mx$ with for $m>0$. Compute the locus of the midpoint of the intersections of lines of the form $y=mx+b$ with the hyperbola, and you should get $y=-x/m$. $\endgroup$
    – amd
    Mar 22, 2019 at 21:18

1 Answer 1


Rectangular hyperbola is given by $x^2-y^2=1$. Consider a diameter given by $y=kx$ (for relevant $k$). In order to describe its conjugate we need to find the midpoints of the chords parallel to $y=kx$. Such a chord is given by $y=kx+n$, so let $(x_1,y_1)$ and $(x_2,y_2)$ be the endpoints of this chord. Note that we are interested only in the midpoint of the chord, i.e. in $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$, hence we don't really need to calculate the coordinates but only $x_1+x_2$ and $y_1+y_2$.

Endpoints of the chord are in the intersection of $x^2-y^2=1$ and $y=kx+n$. If we solve it for $x$, we get $x^2-(kx+n)^2=1$, i.e. $(1-k^2)x^2-2knx-n^2-1=0$. The solutions of this equation are $x_1$ and $x_2$, so by Vieta's formulae we have $x_1+x_2=\frac{2kn}{1-k^2}$.

If we solve the system for $y$, we have $(\frac{y-n}{k})^2-y^2=1$, i.e. $(y-n)^2-k^2y^2=k^2$, and we get $(1-k^2)y^2-2ny+n^2-k^2=0$. As before, $y_1+y_2=\frac{2n}{1-k^2}$.

So the midpoints of the chords are $\{(\frac{kn}{1-k^2},\frac{n}{1-k^2})\mid n\in\mathbb R\}$. This is obviously a line given by $y=\frac{1}{k}x$, which is obviously symmetric to $y=kx$ with respect to an asymptote.

  • $\begingroup$ I was initially seeking some resources, but your proof has helped me grasp the idea. What texts can you recommend that cover similar themes in analytic geometry? (Perhaps I'm merely asking what resource(s) you use(d) ). $\endgroup$
    – E.Nole
    Mar 23, 2019 at 15:47
  • $\begingroup$ @E.Nole I didn't use any resource, this is simply a calculation. To be honest, I don't really know much about books in analytic geometry, so I can't recommend any, but I have a feeling that you can find more on these topics online than in books. $\endgroup$
    – SMM
    Mar 23, 2019 at 19:54
  • $\begingroup$ It was clear to me that you formulated the proof yourself. I meant resources that you personally used to gain the necessary knowledge to come up with the proof. $\endgroup$
    – E.Nole
    Mar 28, 2019 at 19:15

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