The Golden Ellipse: Generation of an Ellipse from a Circle

C o n c e p t s

This topic requires familiarity with the following concepts:

  • Conic equations
  • Differentiation
  • Integration
  • Rotating graphs
  • Golden ratio

The Golden Ellipse was a discovery I made when I read about a curve called the Witch of Agnesi. Using a similar approach, I set out to find the curve with the following property: starting with a circle and a line tangent to it, this curve would define all points that are equidistant from the circle to the line perpendicularly.

The result I obtained was very interesting because the curve was that of a rotated ellipse. I will show you how to derive this ellipse and review some properties.

Defining the Golden Ellipse

The Golden Ellipse

In the figure on the right, the ellipse is generated from the circle by defining a locus of all points P(x, y) such that the distance of a point on the line y = 2R is equal to the distance of a point on the circle.

The curve in green is the circle that generated the ellipse. A point on the curve is equidistant from the line y = 2R and a point on the circle directly to its left horizontally (not necessarily the closest point on the circle).


The general form of the Golden Ellipse is given by: ${{x}^{2}}+2xy+2{{y}^{2}}-4Rx-6Ry+4{{R}^{2}}=0$. The graphing form of the ellipse is $y=\frac{1}{2}(3R-x\pm\sqrt{R^2-2Rx-x^2})$

Later on, I will reveal why I have called it the Golden Ellipse.

Deriving the Equation

Deriving the equation of the ellipse is quite simple. First, note that the vertical distance from the line to point y on the curve (which we know is an ellipse) is 2R - y. This y is equal to the y on the circle. Now, point x on the curve is equal to 2R - y - x, where the second x is the point on the circle. The point x on the circle can be expressed in terms of y which is the same for both the circle and the curve. The equation of the circle is x2 + (y - R)2 = R2. A point on this circle in terms of y is given by $\left( \pm \sqrt{{{R}^{2}}-{{(y-R)}^{2}}},y \right)$. Equating the two distances, we have the equation of the curve:

(i) $2R-y=x\pm \sqrt{{{R}^{2}}-{{(y-R)}^{2}}}$

(ii) ${{(x+y-2R)}^{2}}={{\left( \pm \sqrt{{{R}^{2}}-{{(y-R)}^{2}}} \right)}^{2}}$

(iii) ${{x}^{2}}+{{y}^{2}}+4{{R}^{2}}+2xy-4Rx-4Ry={{R}^{2}}-{{y}^{2}}+2Ry-{{R}^{2}}$

Simplifying (iii) gives us the equation of the ellipse.

(iv) ${{x}^{2}}+2xy+2{{y}^{2}}-4Rx-6Ry+4{{R}^{2}}=0$

Area of the Golden Ellipse

We wonder about the area enclosed by the ellipse. It's quite a surprise that the area of the ellipse is equal to the area of the circle that generated the ellipse! To determine the area, we first find the upper and lower limits of integration by setting $\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}$ equal to zero and solving for x. The x values come out to be $x=R\pm \sqrt{2}R$. Therefore, the area enclosed by the ellipse can be found by the following integral:

(i) $\int\limits_{R-\sqrt{2}R}^{R+\sqrt{2}R}{\text{ }\frac{3R-x+\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}}{2}-\frac{3R-x-\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}}{2}dx}=\int\limits_{R-\sqrt{2}R}^{R+\sqrt{2}R}{\text{ }\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}dx}$

We make the following u-substitution to obtain a friendlier integral: $u=\sqrt{2}R\sin \theta +R$ and $du=\sqrt{2}R\cos \theta \text{ }d\theta$. The limits of integration also simplify to –π/2 and π/2. Therefore, we have:

(ii) $\sqrt{2}R\int\limits_{-{\pi }/{2}\;}^{{\pi }/{2}\;}{\text{ }\sqrt{2{{R}^{2}}-2{{R}^{2}}{{\sin }^{2}}\theta }\cdot \cos \theta \text{ }d\theta }=2{{R}^{2}}\int\limits_{-{\pi }/{2}\;}^{{\pi }/{2}\;}{\text{ }{{\cos }^{2}}\theta \text{ }d\theta }=2{{R}^{2}}\left( \frac{\pi }{2} \right)=\pi{{R}^{2}}$.

The area of the Golden Ellipse is equal to the area of the circle that generated it.

Another method of coming to this conclusion is to realize that the integral $\int\limits_{R-\sqrt{2}R}^{R+\sqrt{2}R}{\text{ }\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}dx}$ represents half the area of the circle $y=\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}$ or ${{(x-R)}^{2}}+{{y}^{2}}=2{{R}^{2}}$. The area of this circle is 2R2π.

Standard from of the Golden Ellipse

The Golden Ellipse

The figure shows the circle in blue used to create the Golden Ellipse in orange. The Golden Ellipse has been rotated by an angle of $\alpha =\arctan \frac{1}{\phi }\approx 31.71{}^\circ$, where Φ is the Golden Ratio. The result is the ellipse in green.

The major axis of the ellipse creates an acute angle of $\alpha =\frac{1}{2}\arctan 2=\arctan \frac{1}{\phi}$, with respect to the negative x-axis, where Φ is the Golden Ratio. Thus, we call this ellipse the Golden Ellipse. The equation of the ellipse in the xy-plane of the circle after it is rotated by α degrees can be determined in general form to be:

(i) $\left( \frac{3-\sqrt{5}}{2} \right){{x}^{2}}+\left( \frac{3+\sqrt{5}}{2} \right){{y}^{2}}-2\sqrt{\frac{65-29\sqrt{5}}{10}}Rx-2\sqrt{\frac{65+29\sqrt{5}}{10}}Ry+4{{R}^{2}}=0$

which can be rewritten as:

$\left( 1-\frac{1}{\phi } \right){{x}^{2}}+(1+\phi ){{y}^{2}}-2\sqrt{\frac{18}{5}+\frac{29}{5}\phi}Rx-2\sqrt{\frac{18}{5}-\frac{29}{5}\cdot\frac{1}{\phi}}Ry+4{{R}^{2}}=0$

In standard form, the equation is:

(ii) $\frac{{{\left( x-\sqrt{\frac{5-2\sqrt{5}}{5}}R \right)}^{2}}}{\frac{3+\sqrt{5}}{2}{{R}^{2}}}+\frac{{{\left( y-\sqrt{\frac{5+2\sqrt{5}}{5}}R \right)}^{2}}}{\frac{3-\sqrt{5}}{2}{{R}^{2}}}=1$ or

$\frac{{{\left( x-\frac{\tan \left( {\pi }/{5}\; \right)}{\sqrt{5}}R \right)}^{2}}}{(1+\phi ){{R}^{2}}}+\frac{{{\left( y-\frac{\tan \left( {2\pi }/{5}\; \right)}{\sqrt{5}}R \right)}^{2}}}{\left( 1-\frac{1}{\phi } \right){{R}^{2}}}=1$

Equation of the rotated ellipse in graphing form:

(iii) $y=\sqrt{\frac{5+2\sqrt{5}}{5}}R\pm R\sqrt{\frac{3-\sqrt{5}}{2}-\left( \frac{7-3\sqrt{5}}{2} \right){{\left( x-\sqrt{\frac{5-2\sqrt{5}}{5}}R \right)}^{2}}}$ or

$y=\frac{\tan \frac{2\pi }{5}}{\sqrt{5}}R\pm R\sqrt{1-\frac{1}{\phi }-\left( \frac{2\phi -3}{\phi } \right){{\left( x-\frac{\tan \frac{\pi }{5}}{\sqrt{5}}R \right)}^{2}}}$

The area of an ellipse is given by abπ. For this ellipse, the area is $\left( \sqrt{\frac{3+\sqrt{5}}{2}}R \right)\left( \sqrt{\frac{3-\sqrt{5}}{2}}R \right)\pi =\sqrt{\frac{4}{4}}\pi {{R}^{2}}=\pi {{R}^{2}}$, which the area of the circle also.

The center of the rotated ellipse is approximately (0.324919696232906∙R, 1.376381920471174∙R).

For both ellipses, the length of the major axis is $2a=2\sqrt{\frac{3+\sqrt{5}}{2}}R$, and the length of the minor axis is $2b=2\sqrt{\frac{3-\sqrt{5}}{2}}R$.

The ellipse is not tangent to the line y = 2R, although is gets very close to the line. The closest point on the ellipse to the line is y ≈ 1.9944159092210683864∙R.

The center of the Golden Ellipse is (R, R). If we translate the ellipse so that its center lies on the origin, then the equation of the ellipse becomes: ${{x}^{2}}+2xy+2{{y}^{2}}-{{R}^{2}}=0$, a much simple equation. The graphing form is $y=\frac{1}{2}\left( -x\pm \sqrt{2{{R}^{2}}-{{x}^{2}}} \right)$. The area of this ellipse, again, can be calculated to be πR2. The equation of this ellipse in the xy-plane is given by: $\frac{{{x}^{2}}}{\frac{3+\sqrt{5}}{2}{{R}^{2}}}+\frac{{{y}^{2}}}{\frac{3-\sqrt{5}}{2}{{R}^{2}}}=1$.

Common Points

Note that the Golden Ellipse and the circle have only two common points. The first common point is trivial—it occurs at (0, 2R). The second common point involves a litte bit of math but the answer is simple. Since we know x = 0 is one of the roots, we expect to solve a polynomial with the constant term being 0. We begin by setting the equations equal:

(i) $3R-x\pm \sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}=2R\pm \sqrt{{{R}^{2}}-{{x}^{2}}}$

(ii) $R-x=\pm 2\sqrt{{{R}^{2}}-{{x}^{2}}}\mp \sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}$

Squaring both sides of (ii):

(iii) ${{R}^{2}}-2Rx+{{x}^{2}}=4({{R}^{2}}-{{x}^{2}})-4\sqrt{{{R}^{4}}+2{{R}^{3}}x-{{R}^{2}}{{x}^{2}}-{{R}^{2}}{{x}^{2}}-2R{{x}^{3}}+{{x}^{4}}}+{{R}^{2}}+2Rx-{{x}^{2}}$

(iv) $2{{R}^{2}}+2Rx-3{{x}^{2}}=2\sqrt{{{x}^{4}}-2R{{x}^{3}}-2{{R}^{2}}{{x}^{2}}+2{{R}^{3}}x+{{R}^{4}}}$

Squaring both sides again:

(v) $9{{x}^{4}}-12R{{x}^{3}}-8{{R}^{2}}{{x}^{2}}+8{{R}^{3}}x+4{{R}^{4}}=4{{x}^{4}}-8R{{x}^{3}}-8{{R}^{2}}{{x}^{2}}+8{{R}^{3}}x+4{{R}^{4}}$

(vi) $5{{x}^{4}}-4R{{x}^{3}}=0$

The solutions of (vi) are $x=0,\text{ }\frac{4R}{5}$.

To find y, we substitute x = 4R/5 to both equations and find the common y produced by the two equations.

Substitution into the circle equation: $y=R\pm \sqrt{{{R}^{2}}-{{\left( \frac{4R}{5} \right)}^{2}}}=R\pm \frac{3R}{5}$. Thus, $y=\frac{8R}{5},\text{ }\frac{2R}{5}$.

Substitution into the ellipse equation: $2y=3R-\frac{4R}{5}\pm \sqrt{{{R}^{2}}+\frac{8{{R}^{2}}}{5}-\frac{16{{R}^{2}}}{25}}=\frac{11R}{5}\pm \frac{7R}{5}$. Thus, $y=\frac{1}{2}\cdot \frac{18R}{5}=\frac{9R}{5}\text{ or }\frac{1}{2}\cdot \frac{4R}{5}=\frac{2R}{5}$.

The common points of the circle and the Golden Ellipse are (0, 2R) and $\left( \frac{4R}{5},\frac{2R}{5} \right)$.

Other Notes

If the ratio of the segments was e rather than 1, then the equation of the curve is:

(i) ${{x}^{2}}+2exy+({{e}^{2}}+1){{y}^{2}}-4eRx-(4{{e}^{2}}R+2R)y+4{{e}^{2}}{{R}^{2}}=0$.

In graphing form, the equation in (i) is: $y=\frac{2{{e}^{2}}R+R-ex\pm \sqrt{{{R}^{2}}+2Re(4{{e}^{2}}-3)x-{{x}^{2}}}}{1+{{e}^{2}}}$.

The equation is the equation of an ellipse and independent of the value of e. This is because the discriminant is always less than 0: ${{B}^{2}}-4AC=4{{e}^{2}}-4({{e}^{2}}+1)=-4$ < 0.