Analytic Geometry

Ellipses

Parts of an Ellipse

An ellipse is a conic section defined by two fixed points. The sum of the distances between these two fixed points and any point on the ellipse is constant.

A focus is a fixed point used to generate a conic section. An ellipse is the set of points such that the sum of the distances from two fixed points, or foci, remains the same. It has a center and passes through two points called vertices and two points called co-vertices. A vertex of an ellipse is one of two points on the ellipse that are endpoints of the major axis. A co-vertex is one of two points on an ellipse that are endpoints of the minor axis.

Properties of an ellipse include:

  • The major axis is the segment that passes through the foci with endpoints on the ellipse. The endpoints are the vertices of the ellipse.
  • The minor axis is the segment with endpoints on an ellipse that is perpendicular to the major axis. The endpoints are the co-vertices of the ellipse.
  • The length of the major axis is always greater than or equal to the length of the minor axis.
  • The major and minor axes intersect at the center of the ellipse.
  • If the lengths of the major and minor axes are equal, then the ellipse is a circle with both foci at the center.
  • The major and minor axes each act as an axis of symmetry for the ellipse, meaning that they divide the figure into two halves that are mirror images.
  • The distance is the same from the center to each focus.
An ellipse is a conic section. Its major axis is equal to or greater than the minor axis. The major and minor axes intersect at the center of the ellipse. The distance from the center and focus are the same.
The equation of an ellipse can be written in terms of its center, vertices, and co-vertices. An ellipse can have a major axis that is vertical or horizontal. For an ellipse in standard form, if the denominator of the xx-term is greater than the denominator of the yy-term, then the major axis is horizontal. If the denominator of the yy-term is greater, then the major axis is vertical.

Equation of an Ellipse

Horizontal Major Axis Centered at the Origin Vertical Major Axis Centered at the Origin
For real numbers aa and bb, where a>ba\gt b, the equation is:
x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
For real numbers aa and bb where a>ba>b, the equation is:
x2b2+y2a2=1\frac{x^2}{b^2}+\frac{y^2}{a^2}=1
The coordinates of the vertices are (a,0)(-a,0) and (a,0)(a,0) . The coordinates of the vertices are (0,a)(0, -a) and (0,a)(0,a) .
The coordinates of the co-vertices are (0,b)(0,-b) and (0,b)(0,b) . The coordinates of the co-vertices are (b,0)(-b,0) and (b,0)(b,0) .
The coordinates of the foci are (c,0)(c,0) and (c,0)(-c,0), where:
b2=a2c2b^2=a^2-c^2
The coordinates of the foci are (0,c)(0,c) and (0,c)(0,-c), where:
b2=a2c2b^2=a^2-c^2

The standard form for an equation of an ellipse with center (h,k)(h, k) is written using (xh)(x-h) in place of xx and (yk)(y-k) in place of yy.

Horizontal major axis:
(xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1
Vertical major axis:
(xh)2b2+(yk)2a2=1\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1

Graphing Ellipses

An ellipse with a given equation can be graphed in the coordinate plane by locating the center, vertices, and co-vertices.

An ellipse can have its center at any point on the coordinate plane, and its major axis may be horizontal or vertical. To graph an ellipse given in standard form:

1. Use the form to determine whether the major axis is horizontal or vertical. For real numbers aa and bb where a>ba>b:

  • The major axis is horizontal if the equation has the form:
(xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1
  • The major axis is vertical if the equation has the form:
(xh)2b2+(yk)2a2=1\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1
2. Use the equation to find the center, (h,k)(h,k) .

3. From the center, use the values of aa and bb to locate the vertices and co-vertices. Then sketch the curve through these points.

  • If the major axis is horizontal, the vertices are aa units to the left and right of the center. The co-vertices are bb units above and below the center.
  • If the major axis is vertical, the vertices are aa units above and below the center. The co-vertices are bb units to the left and right of the center.

4. To locate the foci, solve the equation c2=a2b2c^2=a^2-b^2 to find the value of cc.

  • If the major axis is horizontal, the foci are cc units to the left and right of the center.
  • If the major axis is vertical, the foci are cc units above and below the center.

Ellipses in the Coordinate Plane

Horizontal Major Axis Vertical Major Axis
Equation
x236+y24=1\frac{x^2}{36}+\frac{y^2}{4}=1
x216+y225=1\frac{x^2}{16}+\frac{y^2}{25}=1
Center
(0,0)(0, 0)
(0,0)(0, 0)
Vertices (6,0)(-6, 0) and (6,0)(6, 0) (0,5)(0, 5) and (0,5)(0, -5)
Co-vertices (0,2)(0, 2) and (0,2)(0, -2) (4,0)(-4, 0) and (4,0)(4, 0)
Foci (42,0)(-4\sqrt{2},0) and (42,0)(4\sqrt{2},0) (0,3)(0,3) and (0,3)(0,-3)
Graph

Step-By-Step Example
Graphing an Ellipse in Standard Form
Graph the ellipse with the equation:
(x1)29+(y+2)225=1\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}=1
Step 1
Compare the equation to the standard form for the equation of an ellipse. The denominator of the fraction containing yy is greater than the denominator of the fraction containing xx. So, the standard form is:
(xh)2b2+(yk)2a2=1\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1
This means that the major axis is vertical.
(x1)29+(y+2)225=1\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}=1
The values of hh, kk, b2b^{2}, and a2a^{2} are:
h=1k=2b2=9a2=25\begin{aligned}h&=1 \\k&=-2 \\b^2 &= 9 \\a^2&=25 \end{aligned}
Step 2

Determine the center of the ellipse and the lengths of aa and bb.

Since h=1h=1 and k=2k=-2, the center is (1,2)(1,-2) .

For the length of bb:
b2=9b=9=3\begin{aligned}b^{2}&=9\\b&=\sqrt{9}\\&=3\end{aligned}
For the length of aa:
a2=25a=25=5\begin{aligned}a^{2}&=25\\a&=\sqrt{25}\\&=5\end{aligned}
Step 3

Determine the vertices and co-vertices. The vertices are both 5 units from the center. The major axis is vertical, so the vertices are above and below the center.

Vertices:
(1,2+5)=(1,3)(1,25)=(1,7)\begin{aligned}(1, -2+5)&=(1,3) \\(1, -2-5)&=(1,-7) \end{aligned}
The minor axis is horizontal, and the co-vertices are both 3 units from the center. Co-vertices:
(1+3,2)=(4,2)(13,2)=(2,2)\begin{aligned}(1+3, -2)&=(4,-2)\\(1-3, -2)&=(-2,-2) \end{aligned}
Step 4

Identify the foci.

To locate the foci, use this equation:
c2=a2b2c^2=a^2-b^2
Then determine the value of cc by using the values a=5a=5 and b=3b=3 from Step 2.
c2=a2b2=5232=259=16c=±4\begin{aligned}c^2&=a^2-b^2\\&=5^2-3^2\\&=25-9\\&=16\\c&=\pm4\end{aligned}
The major axis is vertical. So the foci are cc units above and below the center:
(1,24)=(1,6)(1,2+4)=(1,2)\begin{aligned}(1, -2-4)&=(1,-6)\\(1, -2+4)&=(1,2)\end{aligned}
Solution

By Hand: Plot the center, vertices, co-vertices, and foci. Draw a smooth curve through the vertices and co-vertices.

On a Graphing Calculator: Solve for yy:
(x1)29+(y+2)225=1y=±2525(x1)292\begin{aligned}\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}&=1\\y&=\pm\sqrt{25-25\frac{(x-1)^2}{9}}-2\end{aligned}
Graph both semi-ellipses:
Y1=2525(x1)292Y2=2525(x1)292\begin{aligned}Y_1&=\sqrt{25-25\frac{(x-1)^2}{9}}-2 \\ Y_2&=-\sqrt{25-25\frac{(x-1)^2}{9}}-2\end{aligned}
Step-By-Step Example
Graphing an Ellipse by Completing the Square
Graph the ellipse:
4x2+9y2+24x72y+144=04x^2+9y^2+24x-72y+144=0
Step 1
Reorder the terms so that like terms are together and move the constant term to the right side of the equation.
4x2+9y2+24x72y+144=04x2+24x+9y272y=144\begin{aligned}4x^2+9y^2+24x-72y+144&=0\\4x^2+24x+9y^2-72y&=-144\end{aligned}
Step 2
Factor the xx-terms and the yy-terms by grouping so that the coefficient of both squared terms is 1.
4(x2+6x)+9(y28y)=1444(x^2+6x)+9(y^2-8y)=-144
Step 3
Identify the terms that need to be added to complete the square for the xx-terms and the yy-terms.
4(x2+6x+    )+9(y28y+    )=144+4(    )+9(    )4({\color{#c42126} x^2+6x+\underline{\ \ \ \ }})+ 9({\color{#0047af} y^2-8y+\underline{\ \ \ \ }})=-144+4({\color{#c42126} \underline{\ \ \ \ }})+9({\color{#0047af} \underline{\ \ \ \ }})
The missing term for each variable is a number that will make each expression a perfect square. To find this number, divide the coefficient of the middle term by 2 and square the result. For the xx-terms:
(62)2=9\left(\frac{6}{2}\right)^2=9
For the yy-terms:
(82)2=16\left(\frac{-8}{2}\right)^2=16
Step 4

Complete the square for each variable, adding the value to both sides of the equation. Then, factor the expressions on the left side into perfect squares and simplify the right side to write the equation in standard form.

When adding to both sides, remember to account for the coefficients by multiplying the amount added to the xx-terms by 4 and to the yy-terms by 9.
4(x2+6x+    )+9(y28y+    )=144+4(    )+9(    )4(x2+6x+9)+9(y28y+16)=144+4(9)+9(16)4(x+3)2+9(y4)2=36\begin{aligned}4({\color{#c42126} x^2+6x+\underline{\ \ \ \ }})+ 9({\color{#0047af} y^2-8y+\underline{\ \ \ \ }})&=-144+4({\color{#c42126} \underline{\ \ \ \ }})+9({\color{#0047af} \underline{\ \ \ \ }})\\4({\color{#c42126} x^2+6x+9})+9({\color{#0047af} y^2-8y+16})&=-144+4({\color{#c42126} 9})+9({\color{#0047af} 16})\\4({\color{#c42126} x+3})^2+9({\color{#0047af} y-4})^2&=36\end{aligned}
Step 5
Divide both sides of the equation by 36 to write the equation in standard from, with the right side equal to 1.
4(x+3)2+9(y4)2=364(x+3)236+9(y4)236=1(x+3)29+(y4)24=1\begin{aligned}4(x+3)^2+9(y-4)^2&=36\\\frac{4(x+3)^2}{36}+\frac{9(y-4)^2}{36}&=1\\\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}&=1\end{aligned}
Step 6
Identify the center, vertices, and co-vertices.
(x+3)29+(y4)24=1\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}=1
The center of the ellipse is at (3,4)(-3, 4). Since 9>49>4, the major axis is horizontal. Identify the values of aa and bb.
a=9=3b=4=2\begin{aligned}a=\sqrt{9}=3 \\ b=\sqrt{4}=2\end{aligned}
The vertices are 3 units left and right of the center:
(33,4)=(6,4)(3+3,4)=(0,4)\begin{aligned}(-3-3, 4)=(-6,4) \\ (-3+3, 4)=(0,4)\end{aligned}
The co-vertices are 2 units above and below the center:
(3,4+2)=(3,6)(3,42)=(3,2)\begin{aligned}(-3, 4+2)=(-3,6)\\ (-3, 4-2)=(-3,2)\end{aligned}
Step 7

Identify the foci.

To locate the foci, solve the equation c2=a2b2c^2=a^2-b^2 to find the value of cc. Use the values a=3a=3 and b=2b=2 from Step 6.
c2=a2b2=3222=94=5c=±5\begin{aligned}c^2&=a^2-b^2\\&=3^2-2^2\\&=9-4\\&=5\\c&=\pm\sqrt{5}\end{aligned}
The major axis is horizontal. So, the foci are cc units to the left and right of the center: (35,4)(-3-\sqrt{5}, 4) and (3+5,4)(-3+\sqrt{5}, 4) , or about (5.2,4)(-5.2, 4) and (0.8,4)(-0.8, 4) .
Solution

By Hand: Plot the center, vertices, co-vertices, and foci. Draw a smooth curve through the vertices and co-vertices.

On a Graphing Calculator: Solve for yy:
(x+3)29+(y4)24=1y=±44(x+3)29+4\begin{aligned}\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}&=1\\y&=\pm\sqrt{4-\frac{4(x+3)^2}{9}}+4\end{aligned}
Graph both semi-ellipses:
Y1=44(x+3)29+4Y2=44(x+3)29+4\begin{aligned}Y_1&=\sqrt{4-\frac{4(x+3)^2}{9}}+4 \\ Y_2&=-\sqrt{4-\frac{4(x+3)^2}{9}}+4\end{aligned}