I've been taught that a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has it's conjugate hyperbola simply as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$

This also means that the conjugate hyperbola's eccentricity and foci distances are different from the original hyperbola.

The standard hyperbola's eccentricity is given as $\space e= \sqrt{1+\frac{b^2}{a^2}}$

And so, the conjugate hyperbola's eccentricity is (also given by formula $\space \frac{1}{e1^2}+\frac{1}{e2^2}=1$ where e1 and e2 are eccentricities of original hyperbola and it's conjugate) $\space \space e2 = \sqrt{1+ \frac{a^2}{b^2}}$

Now, I was taught that a hyperbola and it's conjugate hyperbola's foci are concyclic, and form the vertices of a square. This is exactly where I get my doubt. I tried plotting some random hyperbola on a graph and then plotted it's conjugate and then tried to see if the foci really are concyclic, and here's what I got.

Random Hyperbola

I made the circle pass through the foci of the standard hyperbola, with P and Q being the foci of the original hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and R and S being the foci of the conjugate. I found out the foci distance by the formula

c (length of foci from origin) = $ae$ for standard hyperbola

or $c = be$ for the conjugate hyperbola

As we see here, the foci aren't forming any square, nor are concyclic.

I request advice to know where I'm missing the mark.

  • $\begingroup$ @Natru: What "graphical explanation" do you expect? OP simply mis-calculated somewhere; without intermediate calculations, we can only guess about the error. ... OP knows the distances to the foci are $c_1=ae_1$ and $c_2=be_2$, where $e_1:=\sqrt{1+\frac{b^2}{a^2}}$ and $e_2:=\sqrt{1+\frac{a^2}{b^2}}$ (ie, $5/4$ and $5/3$). OP could have reduced algebraically to get $c_i=\sqrt{a^2+b^2}$ (ie, $5$) for both cases, but didn't. Instead, OP calculated $c_1=20/3$ and $c_2=15/4$. OP may have inadvertently swapped the $e_i$ ... or something else. No "graphical explanation" will solve that mystery. $\endgroup$
    – Blue
    Commented Jan 22, 2022 at 5:23
  • $\begingroup$ I had made a miscalculation myself too, they do form a square. But a proof on why the focii are concyclic would be appreciated. $\endgroup$
    – Natru
    Commented Jan 23, 2022 at 11:21

3 Answers 3


It's not clear to me where things are going wrong for you, but you have a lot of correct information.

Ultimately, you are correct that the distance $c$ from center to focus is given by $c = a e$ for the standard hyperbola, and by $c=be$ for the conjugate hyperbola. But be careful: you mean $c=ae_1$ and $c=be_2$, since the eccentricities differ. Notice what those calculations give: $$\begin{align} \text{standard:}\quad c &= a e_1=a\;\sqrt{1+\frac{b^2}{a^2}}= a\;\sqrt{\frac{a^2+b^2}{a^2}}=a\;\frac{\sqrt{a^2+b^2}}{a}=\sqrt{a^2+b^2} \tag{1} \\[6pt] \text{conjugate:}\quad c &= b e_2=b\;\sqrt{1+\frac{a^2}{b^2}}= b\;\sqrt{\frac{b^2+a^2}{b^2}}=b\;\frac{\sqrt{b^2+a^2}}{b}=\sqrt{b^2+a^2} \tag{2} \end{align}$$ Since $(1)$ and $(2)$ match, the standard and conjugate foci are all equidistant from the origin; thus, they are the vertices of a square, and they must lie on a common circle. $\square$

So, I would guess that something is wrong with your Desmos entries, but I'm not currently in a position to pick through those details. You might consider using GeoGebra, where you can enter the algebraic formulas for the hyperbolas but also construct them geometrically via their foci, and see if the results match.


Hope the Geogebra sketch explains all your questions. Main/conjugate hyperbolae constructed with major/minor axes.

$$(4,5);\, (3,5);$$

The central rectangle should be drawn. Their eccentricities

$$e_1= AC/AG ;\, e_2= AC/AF ;\quad \frac{1}{e_1^2}+\frac{1}{e_2^2}=1\,$$

The concyclic foci $I,H$ and the other two axis mirrored foci are seen on a circle.



Th3 foci of the horizonal Hyperbola are at $H_1,H_2=(\pm ea,0)$ and yhose of bertical ones are $V_1,V_2=(0,\pm bE)$. Here $$e=\frac{\sqrt{a^2+b^2}}{a},~~ E=\frac{\sqrt{a^2+b^2}}{b}$$ $O(0,0)$ being the center the concyclicity of$ H_1,H_2,V_1, V_2$ demands that $$OH_1.OH_2=OV_1.OV_2 \implies a^2e^2=b^2E^2= (a^2+b^2),$$ and this is true.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .