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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.

Figure 1

[Source: Wikipedia]

Ellipse

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

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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

close solution
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

1601 video

video by PatrickJMT

close solution

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