17Calculus - Ellipses in Rectangular Coordinates
An ellipse is formed when a plane intersects a cone not parallel to one of sides of the cone and not parallel to the axis of the cone. A circle is a special form of an ellipse where the plane is perpendicular to the axis of the cone.
On this page, we discuss ellipses in rectangular coordinates. For a discussion of ellipses in polar form, see this separate page.
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Definition of an Ellipse
In the next section, we show two different ways to get an ellipse. There are actually three ways but we will stick with two for now. One way is to slice a cone with a plane. The other way you may be more familiar with from high school. You draw an ellipse with a piece of string and two thumbtacks.
So you may be asking why do those two different definitions define the same figure? I mean, they seem so different. Well, here is a great video that explains this. This is absolutely fascinating!
3Blue1Brown - Why slicing a cone gives an ellipse [12min-51secs]
video by 3Blue1Brown |
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Ellipse
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Figure 1 |
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Figure 2 |
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The standard equation for an ellipse is \(\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }\). Figure 2 contains more information than we need right now but it will suffice. The longer axis is called the major axis (in this plot it is horizontal). The shorter axis is called the minor axis. The vertices are located on the ellipse where it crosses the major axis. The foci are also on the major axis, labeled F1 and F2 on this plot.
The major axis is determined by the denominators, \(a^2\) and \(b^2\). The larger value is in the denominator of the major axis, i.e. if \(a > b\) then the major axis is parallel to the x-axis. We need to define a value c where \(c^2=\abs{a^2-b^2}\) which will help us determine the location of the foci.
These tables contain the main attributes of an ellipse. We assume here that \(a > b\). Similar equations exist for \(a < b\).
\(\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }\) | |
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center |
\((h,k)\) |
major axis |
\(y=k\) |
vertices |
\((h \pm a, k), \) \( (h, k \pm b)\) |
foci |
\( (h \pm c,k) \) |
\(c^2=\abs{a^2-b^2}\) | |
eccentricity |
\( e=c/a \) |
\(\displaystyle{ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 }\) | |
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center |
\((h,k)\) |
major axis |
\(x=h\) |
vertices |
\((h, k \pm a), \) \( (h \pm b, k)\) |
foci |
\( (h,k \pm c) \) |
\(c^2=\abs{a^2-b^2}\) | |
eccentricity |
\( e=c/a \) |
Notes |
1. Since the foci are closer to the center than the vertices, it follows that \(c < a\) and therefore \(0 < e < 1\).
2. Notice in the standard form of the equation, both terms are positive. This is how you know the graph is an ellipse and not a hyperbola.
3. In the general form of the equation, \(Ax^2+Bxy+Cy^2+\) \(Dx+Ey+F=0\), \(A > 0\) and \(C > 0\).
4. The eccentricity e is not the same as the irrational constant \(e \approx 2.72\).
Okay, time for some fun videos about ellipses. Here are a couple of videos about playing pool on an elliptical table. They clearly show the relationship between the foci and demonstrate some fun physics at the same time.
Numberphile - Elliptical Pool Table (1) [3min-39secs]
video by Numberphile |
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Numberphile - Elliptical Pool Table (2) [5min-52secs]
video by Numberphile |
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Perimeter of an Ellipse
Did you know that there is no simple, closed form of an equation for the perimeter of an ellipse? Really?! I didn't either until recently. I've taught ellipses in precalc and calc multiple times and I never knew this! So here is the video that opened my eyes to this strange and wonderful fact. At about time index 14:11 he explains that the perimeter is actually calculated using an infinite series. Enjoy.
Stand Up Maths - Why is there no equation for the perimeter of an ellipse? [21min-4secs]
video by Stand Up Maths |
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Practice
Write the standard form of the equation for an ellipse, centered at the origin, vertical major axis of length 8 and minor axis of length 2.
Problem Statement |
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Write the standard form of the equation for an ellipse, centered at the origin, vertical major axis of length 8 and minor axis of length 2.
Solution |
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Write the standard form of the equation for an ellipse, centered at the origin, with x-intercepts at \(\pm 12\) and foci at \((0,\pm 5)\).
Problem Statement |
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Write the standard form of the equation for an ellipse, centered at the origin, with x-intercepts at \(\pm 12\) and foci at \((0,\pm 5)\).
Solution |
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1588 video
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Write the standard form of the equation for an ellipse, centered at the origin, with minor axis of length 6 and foci at \( (\pm 8, 0) \).
Problem Statement |
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Write the standard form of the equation for an ellipse, centered at the origin, with minor axis of length 6 and foci at \( (\pm 8, 0) \).
Solution |
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1589 video
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Find the intercepts of the ellipse \(\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }\).
Problem Statement |
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Find the intercepts of the ellipse \(\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }\).
Solution |
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1590 video
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Graph the ellipse \(\displaystyle{ 1 - \frac{y^2}{16} = x^2 }\).
Problem Statement |
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Graph the ellipse \(\displaystyle{ 1 - \frac{y^2}{16} = x^2 }\).
Solution |
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1591 video
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Find the center and the radius of the circle \(x^2+2x+y^2=4\).
Problem Statement |
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Find the center and the radius of the circle \(x^2+2x+y^2=4\).
Solution |
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1594 video
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Sketch the circle \( x^2 + y^2 = 4x \).
Problem Statement |
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Sketch the circle \( x^2 + y^2 = 4x \).
Solution |
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1595 video
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Sketch the circle \( x^2 + y^2 + 6y = 0 \).
Problem Statement |
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Sketch the circle \( x^2 + y^2 + 6y = 0 \).
Solution |
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1596 video
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Sketch the circle \( x^2 + y^2 + 2x + 2y = 2 \).
Problem Statement |
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Sketch the circle \( x^2 + y^2 + 2x + 2y = 2 \).
Solution |
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1597 video
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Sketch the circle \( x^2 + y^2 + 10x - 20y + 100 = 0\).
Problem Statement |
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Sketch the circle \( x^2 + y^2 + 10x - 20y + 100 = 0\).
Solution |
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1598 video
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Sketch the circle \( 2x^2 + 2y^2 + 2x - 2y = 1 \).
Problem Statement |
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Sketch the circle \( 2x^2 + 2y^2 + 2x - 2y = 1 \).
Solution |
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1599 video
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Sketch the circle \( 9x^2 + 9y^2 - 6x - 12y = 11 \).
Problem Statement |
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Sketch the circle \( 9x^2 + 9y^2 - 6x - 12y = 11 \).
Solution |
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Graph \(\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }\).
Problem Statement |
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Graph \(\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }\).
Solution |
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Topics You Need To Understand For This Page
Related Topics and Links
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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