• Trigonometry > The Golden Ellipse

The Golden Ellipse: Generation of an Ellipse from a Circle

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 to the circle horizontally and to the line perpendicularly.

The result I obtained was 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

In Figure 1, the ellipse is generated from the circle by defining a locus of all points such that the distance of a point on the line y = 2R is equal to the distance of a point on the circle.

We will use the circle in blue to generate the ellipse. On the right half of the circle, the point E₁ on the curve is equidistant from the line y = 2R at point L₁ and to a point on the circle C₁ directly to its left horizontally (not necessarily the closest point on the circle). On the left half of the circle, point E₂ is equidistant to point L₂ on the line and point C₂ on the circle.

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

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. From the definition of our curve, L₁E₁ = E₁C₁.

First, we note that the y value of C₁ (the point on the circle) and E₁ are equal. Therefore, we know the curve’s y value at E₁.

Now, the length of segment L₁E₁ is equal to 2R – y, where y is the y value of point C₁ on the circle. This length is also equal to the length of segment C₁E₁.

Therefore, x value of point E₁ is equal to L₁E₁ + x, where x is equal to the x value of C₁. Therefore, the x value of E₁ is equal to 2R – y + x, where both x and y are the coordinates of the circle.

So far, we have the point E₁ identified based on the coordinates of the circle, and that point is: \(\text{E}_1 = (2R - y + x, y)\).

The equation of the circle is \(x^2 + (y-R)^2 = R^2\). A point on this circle in terms of y is given by \(\left( \pm \sqrt{R^2-(y-R)^2},y \right)\). We can substitute the x coordinate of the circle for x in point E₁ and express E₁ in terms of y:

(i) \(\text{E}_1 = (2R - y \pm\sqrt{R^2 - (y-R)^2}, y)\)

Equation (i) is basically an inverse function that we can solve for y.

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

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

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

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

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

Based on how we defined the ellipse, we can easily determine the center of the ellipse is (R, R).

Standard Form of the Golden Ellipse

The Angle of Rotation

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 equal to \(\frac{1+\sqrt{5}}{2}\). The result is the ellipse in green.

The angle of rotation for a general conic defined by the equation \(Ax^2+Bxy+Cy^2+Dx+Ey+F = 0\) is \(\alpha = \frac{1}{2}\arctan\frac{B}{C-A}\). (Rotation of conics has been covered on the Conic Sections page.) Based on our ellipse equation, A = 1, B = 2, and C = 2. Therefore, the angle of rotation, α, is: \(\alpha = \frac{1}{2}\arctan\frac{2}{2-1} = \frac{1}{2}\arctan 2\).

The arctangent of 2 has a famous connection with the Golden Ratio: \(\arctan 2 = 2\arctan\frac{1}{\phi}\).

The major axis of the ellipse also makes an acute angle α with respect to the negative x-axis. Because the angle of rotation has the Golden Ratio in its formula, we call this ellipse the Golden Ellipse.

The sine and cosine of the angle of rotation can be useful for finding the location of the center and the foci of the ellipse. The cosine value is \(\cos\alpha = \sqrt{\frac{5+\sqrt{5}}{10}}\) and the sine value is \(\sin\alpha = \sqrt{\frac{5-\sqrt{5}}{10}}\).

The Standard Equation

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 (ii) allows us to find the measure of the major and minor axes, or the lengths 2a and 2b.

For both ellipses, the length of the major axis is \(2a = 2R\sqrt{1+\phi} = 2\sqrt{\frac{3+\sqrt{5}}{2}}R\), and the length of the minor axis is \(2b = 2R\sqrt{1-\frac{1}{\phi}} = 2\sqrt{\frac{3-\sqrt{5}}{2}}R\).

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 center of the rotated ellipse is approximately (0.324919696232906∙R, 1.376381920471174∙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. This value can be found by adding the location of the y value of the center of the ellipse with the length of the minor axis: \(\sqrt{\frac{5+2\sqrt{5}}{5}}+\sqrt{\frac{3-\sqrt{5}}{2}} \approx 1.99R\).

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 equation of this ellipse in the xy-plane is given by: \(\frac{{{x}^{2}}}{\frac{3+\sqrt{5}}{2}{{R}^{2}}}+\frac{{{y}^{2}}}{\frac{3-\sqrt{5}}{2}{{R}^{2}}}=1\).

Eccentricity

The values of a and b do not change when rotating or translating the ellipse. Therefore, knowing a and b from the standard form of the equation, we can find c and the eccentricity, e, of the ellipse.

(i) \(c^2 = a^2 - b^2\)

(ii) \(c^2 = R^2\left(\frac{3+\sqrt{5}}{2}\right) - R^2\left(\frac{3-\sqrt{5}}{2}\right)\)

(iii) \(c^2 = R^2\left(\frac{3}{2} + \frac{\sqrt{5}}{2}-\frac{3}{2} + \frac{\sqrt{5}}{2}\right)\)

(iv) \(c^2 = R^2\sqrt{5}\)

(v) \(c = \sqrt[4]{5}R\)

We can find the eccentricity now that we know c.

(i) \(e = \frac{c}{a}\)

(ii) \(e = \frac{\sqrt[4]{5}R}{\sqrt{\frac{3+\sqrt{5}}{2}}R}\)

(iii) \(e = \frac{1}{2}\sqrt{6\sqrt{5}-10} \approx 0.924\)

The eccentricity seems to be high and close to 1 (of course, not enough to look like a parabola).

The Foci

Foci of the Standard Ellipse

Knowing c from equation (v) above, we can find the location of the foci of the standard ellipse since we know a.

The foci, F₁ and F₂ are located at: \(\text{F}_1 = \left(R\sqrt{\frac{5-2\sqrt{5}}{5}} - R\sqrt[4]{5}, R\sqrt{\frac{5+2\sqrt{5}}{5}} \right)\) and \(\text{F}_2 = \left(R\sqrt{\frac{5-2\sqrt{5}}{5}} + R\sqrt[4]{5}, R\sqrt{\frac{5+2\sqrt{5}}{5}} \right)\). The foci are very close to the vertices of the major axis because the eccentricity is high.

You can see the foci of the standard ellipse in Figure 3 below.

If the center of the ellipse was the origin, our foci would simply be \(R\sqrt[4]{5}\) distance away from the origin.

Foci of the Golden Ellipse

To find the foci of the Golden Ellipse, it is easier to use right triangles than rotate the focal points. Figure 3 can help us find the location of the foci.

In Figure 4, the coordinates of O are (R, R). The length of OF₂ is equal to c, which we found above to equal \(\sqrt[4]{5}\). The angle OF₂Q is equal to α, which is the angle of rotation. We know the sine and cosine values of α.

Therefore, to find the horizontal displacement, Δx, of the foci from the center, we can use the cosine value of α: \(\Delta x = R\sqrt[4]{5}\cos\alpha = R\sqrt[4]{5}\sqrt{\frac{5+\sqrt{5}}{10}} = R\sqrt{\frac{\sqrt{5}+1}{2}}\).

The vertical displacement from the center, Δy, is \(\Delta y = R\sqrt[4]{5}\sin\alpha = R\sqrt[4]{5}\sqrt{\frac{5-\sqrt{5}}{10}} = R\sqrt{\frac{\sqrt{5}-1}{2}}\).

Applying the displacements to the center of the Golden Ellipse, our foci are located at: \(\text{F}_1 = \left( R - R\sqrt{\frac{\sqrt{5}+1}{2}}, R + R\sqrt{\frac{\sqrt{5}-1}{2}} \right)\) and \(\text{F}_2 = \left( R + R\sqrt{\frac{\sqrt{5}+1}{2}}, R - R\sqrt{\frac{\sqrt{5}-1}{2}} \right)\)

Critical Points

If the ellipse is in standard form, we can easily find the critical points, namely the vertices. However, in the slanted ellipse, we have some more critical points in addition to the vertices that have been rotated.

In Figure 6, points A and B are the farthest points horizontally and the tangents at these points are vertical lines. Points C and D are the farthest points vertically and the tangents at these points are horizontal lines, which are the x-axis and the line y = 2R.

Points A and B represent the largest and smallest y values, and these are trivial since the highest y value will be 2R when x = 0 and lowest will be 0 when x = 2R. So these points are (0, 2R) and (2R, 0).

Points C and D represent the minimum and maximum x values. These two can be found by solving for x in the rectangular form of the equation: \(y= \frac{3}{2}R -\frac{1}{2}x \pm\frac{1}{2}\sqrt{R^2+2Rx-x^2})\). These points will be when the radicand is 0, so we just need to solve this for 0: \(R^2+2Rx-x^2 = 0\). The solutions to this equation are \(x = R(1\pm\sqrt{2})\).

We can find the y values by substitution. Therefore, the points \(\left(R(1-\sqrt{2}), R\left(1+\frac{\sqrt{2}}{2}\right) \right)\) and \(\left(R(1+\sqrt{2}), R\left(1-\frac{\sqrt{2}}{2}\right) \right)\) are the farthest points horizontally.

To find vertices of the major axis, we can use the same method for finding the foci.

We know that OV₁ = a and angle OV₁Q is equal to α. So, our horizontal displacement, Δx, from the center is: \(\Delta x = a\cos\alpha = R\sqrt{\frac{3+\sqrt{5}}{2}}\sqrt{\frac{5+\sqrt{5}}{10}} = R\sqrt{\frac{5+2\sqrt{5}}{5}}\).

The vertial displacement from the center, Δy, is: \(\Delta y = a\sin\alpha = R\sqrt{\frac{3+\sqrt{5}}{2}}\sqrt{\frac{5-\sqrt{5}}{10}} = R\sqrt{\frac{5+\sqrt{5}}{10}}\).

Our vertices on the major axis are: \(\left( R + R\sqrt{\frac{5+2\sqrt{5}}{5}}, R - R\sqrt{\frac{5+\sqrt{5}}{10}} \right)\) and \(\left( R - R\sqrt{\frac{5+2\sqrt{5}}{5}}, R + R\sqrt{\frac{5+\sqrt{5}}{10}} \right)\).

The vertices of the minor axis, the horizontal and vertical displacements are \(\Delta x = b\sin\alpha = \sqrt{\frac{5-2\sqrt{5}}{5}}\) and \(\Delta y = b\cos\alpha = \sqrt{\frac{5-\sqrt{5}}{10}}\). Note that the sine and cosine got switched because the triangle V₃OR is in a different orientation. The angle of rotation is OV₃R.

The two vertices are: \(\left( R + R\sqrt{\frac{5-2\sqrt{5}}{5}}, R + R\sqrt{\frac{5-\sqrt{5}}{10}} \right)\) and \(\left( R - R\sqrt{\frac{5-2\sqrt{5}}{5}}, R - R\sqrt{\frac{5-\sqrt{5}}{10}} \right)\).

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! Actually, not a big surprise, since a horizontal cross section of the ellipse at any point is also equal to the horizontal cross section of the circle. But we will calculate the area using calculus.

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) \(\displaystyle\int_{R-\sqrt{2}R}^{R+\sqrt{2}R} \frac{3R-x+\sqrt{{R^2}+2Rx-{x^2}}}{2}-\) \(\frac{3R-x-\sqrt{{R^2}+2Rx-{x^2}}}{2}dx=\) \(\int_{R-\sqrt{2}R}^{R+\sqrt{2}R}{\sqrt{{{R}^{2}}+2Rx-{{x}^{2}}}\text{ }dx}\)

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

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

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 \(\int_{R-\sqrt{2}R}^{R+\sqrt{2}R}{\text{ }\sqrt{{R^2}+2Rx-{x^2}}dx}\) 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.

And yet another way to determine the area is from the area of an ellipse formula: 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.

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, \(y=\frac{1}{2}\cdot \frac{18R}{5}=\frac{9R}{5}\) 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 (not the eccentricity) 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: \({{B}^{2}}-4AC=4{{e}^{2}}-4({{e}^{2}}+1)=-4\) is less than 0.

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