\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Ellipses in Rectangular Coordinates

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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.

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

Ellipse

Figure 1

[Source: Wikipedia]

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 \(a < b\).

\(\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }\)

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

\( e=c/a \)

\(\displaystyle{ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 }\)

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

Numberphile - Elliptical Pool Table (2) [5min-52secs]

video by Numberphile

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

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

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

1587 video

video by PatrickJMT

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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

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

1588 video

video by PatrickJMT

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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

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

1589 video

video by PatrickJMT

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Find the intercepts of the ellipse \(\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }\).

Problem Statement

Find the intercepts of the ellipse \(\displaystyle{ \frac{y^2}{100} + \frac{x^2}{121} = 1 }\).

Solution

1590 video

video by PatrickJMT

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Graph the ellipse \(\displaystyle{ 1 - \frac{y^2}{16} = x^2 }\).

Problem Statement

Graph the ellipse \(\displaystyle{ 1 - \frac{y^2}{16} = x^2 }\).

Solution

1591 video

video by PatrickJMT

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Find the center and the radius of the circle \(x^2+2x+y^2=4\).

Problem Statement

Find the center and the radius of the circle \(x^2+2x+y^2=4\).

Solution

1594 video

video by Krista King Math

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Sketch the circle \( x^2 + y^2 = 4x \).

Problem Statement

Sketch the circle \( x^2 + y^2 = 4x \).

Solution

1595 video

video by Krista King Math

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Sketch the circle \( x^2 + y^2 + 6y = 0 \).

Problem Statement

Sketch the circle \( x^2 + y^2 + 6y = 0 \).

Solution

1596 video

video by Krista King Math

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Sketch the circle \( x^2 + y^2 + 2x + 2y = 2 \).

Problem Statement

Sketch the circle \( x^2 + y^2 + 2x + 2y = 2 \).

Solution

1597 video

video by Krista King Math

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Sketch the circle \( x^2 + y^2 + 10x - 20y + 100 = 0\).

Problem Statement

Sketch the circle \( x^2 + y^2 + 10x - 20y + 100 = 0\).

Solution

1598 video

video by Krista King Math

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Sketch the circle \( 2x^2 + 2y^2 + 2x - 2y = 1 \).

Problem Statement

Sketch the circle \( 2x^2 + 2y^2 + 2x - 2y = 1 \).

Solution

1599 video

video by Krista King Math

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Sketch the circle \( 9x^2 + 9y^2 - 6x - 12y = 11 \).

Problem Statement

Sketch the circle \( 9x^2 + 9y^2 - 6x - 12y = 11 \).

Solution

1600 video

video by Krista King Math

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Graph \(\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }\).

Problem Statement

Graph \(\displaystyle{ \frac{x^2}{9} + \frac{y^2}{5} = 1 }\).

Solution

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

Ellipse - Wikipedia

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

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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