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Guias de estudo > College Algebra

Equations of Ellipses

Learning Objectives

  • Identify the foci, vertices, axes, and center of an ellipse
  • Write equations of ellipses centered at the origin
  • Write equations of ellipses not centered at the origin
A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape. Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points (x,y)\left(x,y\right) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x- and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.

Writing Equations of Ellipses Not Centered at the Origin

Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated hh units horizontally and kk units vertically, the center of the ellipse will be (h,k)\left(h,k\right). This translation results in the standard form of the equation we saw previously, with xx replaced by (x−h)\left(x-h\right) and y replaced by (y−k)\left(y-k\right).

A General Note: Standard Forms of the Equation of an Ellipse with Center (h, k)

The standard form of the equation of an ellipse with center (h, k)\left(h,\text{ }k\right) and major axis parallel to the x-axis is

(x−h)2a2+(y−k)2b2=1\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1

where
  • a>ba>b
  • the length of the major axis is 2a2a
  • the coordinates of the vertices are (h±a,k)\left(h\pm a,k\right)
  • the length of the minor axis is 2b2b
  • the coordinates of the co-vertices are (h,k±b)\left(h,k\pm b\right)
  • the coordinates of the foci are (h±c,k)\left(h\pm c,k\right), where c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.
The standard form of the equation of an ellipse with center (h,k)\left(h,k\right) and major axis parallel to the y-axis is

(x−h)2b2+(y−k)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1

where
  • a>ba>b
  • the length of the major axis is 2a2a
  • the coordinates of the vertices are (h,k±a)\left(h,k\pm a\right)
  • the length of the minor axis is 2b2b
  • the coordinates of the co-vertices are (h±b,k)\left(h\pm b,k\right)
  • the coordinates of the foci are (h,k±c)\left(h,k\pm c\right), where c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.
Just as with ellipses centered at the origin, ellipses that are centered at a point (h,k)\left(h,k\right) have vertices, co-vertices, and foci that are related by the equation c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
(a) Horizontal ellipse with center (h,k) (b) Vertical ellipse with center (h,k) (a) Horizontal ellipse with center (h,k)\left(h,k\right) (b) Vertical ellipse with center (h,k)\left(h,k\right)

How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.

  1. Determine whether the major axis is parallel to the x- or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form (x−h)2a2+(y−k)2b2=1\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1.
    2. If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x−h)2b2+(y−k)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1.
  2. Identify the center of the ellipse (h,k)\left(h,k\right) using the midpoint formula and the given coordinates for the vertices.
  3. Find a2{a}^{2} by solving for the length of the major axis, 2a2a, which is the distance between the given vertices.
  4. Find c2{c}^{2} using hh and kk, found in Step 2, along with the given coordinates for the foci.
  5. Solve for b2{b}^{2} using the equation c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.
  6. Substitute the values for h,k,a2h,k,{a}^{2}, and b2{b}^{2} into the standard form of the equation determined in Step 1.

Example: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin

What is the standard form equation of the ellipse that has vertices (−2,−8)\left(-2,-8\right) and (−2,2)\left(-2,\text{2}\right) and foci (−2,−7)\left(-2,-7\right) and (−2,1)?\left(-2,\text{1}\right)?

Answer: The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form

(x−h)2b2+(y−k)2a2=1\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1

First, we identify the center, (h,k)\left(h,k\right). The center is halfway between the vertices, (−2,−8)\left(-2,-8\right) and (−2,2)\left(-2,\text{2}\right). Applying the midpoint formula, we have:

(h,k)=(−2+(−2)2,−8+22) =(−2,−3)\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}

Next, we find a2{a}^{2}. The length of the major axis, 2a2a, is bounded by the vertices. We solve for aa by finding the distance between the y-coordinates of the vertices.

2a=2−(−8)2a=10a=5\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}

So a2=25{a}^{2}=25. Now we find c2{c}^{2}. The foci are given by (h,k±c)\left(h,k\pm c\right). So, (h,k−c)=(−2,−7)\left(h,k-c\right)=\left(-2,-7\right) and (h,k+c)=(−2,1)\left(h,k+c\right)=\left(-2,\text{1}\right). We substitute k=−3k=-3 using either of these points to solve for cc.

k+c=1−3+c=1c=4\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array} So c2=16{c}^{2}=16.

Next, we solve for b2{b}^{2} using the equation c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.

c2=a2−b216=25−b2b2=9\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}

Finally, we substitute the values found for h,k,a2h,k,{a}^{2}, and b2{b}^{2} into the standard form equation for an ellipse:

(x+2)29+(y+3)225=1\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1

Try It

What is the standard form equation of the ellipse that has vertices (−3,3)\left(-3,3\right) and (5,3)\left(5,3\right) and foci (1−23,3)\left(1 - 2\sqrt{3},3\right) and (1+23,3)?\left(1+2\sqrt{3},3\right)?

Answer: (x−1)216+(y−3)24=1\frac{{\left(x - 1\right)}^{2}}{16}+\frac{{\left(y - 3\right)}^{2}}{4}=1

 

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