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学習ガイド > College Algebra CoRequisite Course

Graphs of Ellipses

Learning Outcomes

  • Sketch a graph of an ellipse centered at the origin.
  • Sketch a graph of an ellipse not centered at the origin.
  • Express the equation of an ellipse in standard form given the equation in general form.
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form x2a2+y2b2=1, a>b\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b for horizontal ellipses and x2b2+y2a2=1, a>b\dfrac{{x}^{2}}{{b}^{2}}+\dfrac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b for vertical ellipses.

How To: Given the standard form of an equation for an ellipse centered at (0,0)\left(0,0\right), sketch the graph.

  • Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
    • If the equation is in the form x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1, where a>ba>b, then
      • the major axis is the x-axis
      • the coordinates of the vertices are (±a,0)\left(\pm a,0\right)
      • the coordinates of the co-vertices are (0,±b)\left(0,\pm b\right)
      • the coordinates of the foci are (±c,0)\left(\pm c,0\right)
    • If the equation is in the form x2b2+y2a2=1\dfrac{{x}^{2}}{{b}^{2}}+\dfrac{{y}^{2}}{{a}^{2}}=1, where a>ba>b, then
      • the major axis is the y-axis
      • the coordinates of the vertices are (0,±a)\left(0,\pm a\right)
      • the coordinates of the co-vertices are (±b,0)\left(\pm b,0\right)
      • the coordinates of the foci are (0,±c)\left(0,\pm c\right)
  • Solve for cc using the equation c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.
  • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

Recall the square root property

With the x2{x}^{2} term isolated, the square root propty states that:

if x2=k,then x=±k\text{if }{x}^{2}=k,\text{then }x=\pm \sqrt{k}

Don't forget to take the plus or minus!

Example: Graphing an Ellipse Centered at the Origin

Graph the ellipse given by the equation, x29+y225=1\dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{25}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: First, we determine the position of the major axis. Because 25>925>9, the major axis is on the y-axis. Therefore, the equation is in the form x2b2+y2a2=1\dfrac{{x}^{2}}{{b}^{2}}+\dfrac{{y}^{2}}{{a}^{2}}=1, where b2=9{b}^{2}=9 and a2=25{a}^{2}=25. It follows that:

  • the center of the ellipse is (0,0)\left(0,0\right)
  • the coordinates of the vertices are (0,±a)=(0,±25)=(0,±5)\left(0,\pm a\right)=\left(0,\pm \sqrt{25}\right)=\left(0,\pm 5\right)
  • the coordinates of the co-vertices are (±b,0)=(±9,0)=(±3,0)\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)
  • the coordinates of the foci are (0,±c)\left(0,\pm c\right), where c2=a2−b2{c}^{2}={a}^{2}-{b}^{2} Solving for cc, we have:

\begin{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{25 - 9} \\ &=\pm \sqrt{16} \\ &=\pm 4 \end{align}

Therefore, the coordinates of the foci are (0,±4)\left(0,\pm 4\right). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation x236+y24=1\dfrac{{x}^{2}}{36}+\dfrac{{y}^{2}}{4}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: center: (0,0)\left(0,0\right); vertices: (±6,0)\left(\pm 6,0\right); co-vertices: (0,±2)\left(0,\pm 2\right); foci: (±42,0)\left(\pm 4\sqrt{2},0\right)

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Example: Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form

Graph the ellipse given by the equation 4x2+25y2=1004{x}^{2}+25{y}^{2}=100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Answer: First, use algebra to rewrite the equation in standard form.

4x2+25y2=1004x2100+25y2100=100100x225+y24=1\begin{gathered} 4{x}^{2}+25{y}^{2}=100 \\[1.5mm] \dfrac{4{x}^{2}}{100}+\dfrac{25{y}^{2}}{100}=\dfrac{100}{100} \\[1.5mm] \dfrac{{x}^{2}}{25}+\dfrac{{y}^{2}}{4}=1 \end{gathered}

Next, we determine the position of the major axis. Because 25>425>4, the major axis is on the x-axis. Therefore, the equation is in the form x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1, where a2=25{a}^{2}=25 and b2=4{b}^{2}=4. It follows that:
  • the center of the ellipse is (0,0)\left(0,0\right)
  • the coordinates of the vertices are (±a,0)=(±25,0)=(±5,0)\left(\pm a,0\right)=\left(\pm \sqrt{25},0\right)=\left(\pm 5,0\right)
  • the coordinates of the co-vertices are (0,±b)=(0,±4)=(0,±2)\left(0,\pm b\right)=\left(0,\pm \sqrt{4}\right)=\left(0,\pm 2\right)
  • the coordinates of the foci are (±c,0)\left(\pm c,0\right), where c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}. Solving for cc, we have:

\begin{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{25 - 4} \\ &=\pm \sqrt{21} \end{align}

Therefore the coordinates of the foci are (±21,0)\left(\pm \sqrt{21},0\right). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation 49x2+16y2=78449{x}^{2}+16{y}^{2}=784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Answer: Standard form: x216+y249=1\dfrac{{x}^{2}}{16}+\dfrac{{y}^{2}}{49}=1; center: (0,0)\left(0,0\right); vertices: (0,±7)\left(0,\pm 7\right); co-vertices: (±4,0)\left(\pm 4,0\right); foci: (0,±33)\left(0,\pm \sqrt{33}\right)

Graphing Ellipses Not Centered at the Origin

When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h,k)\left(h,k\right), we use the standard forms (x−h)2a2+(y−k)2b2=1, a>b\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{ }a>b for horizontal ellipses and (x−h)2b2+(y−k)2a2=1, a>b\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{ }a>b for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.

How To: Given the standard form of an equation for an ellipse centered at (h,k)\left(h,k\right), sketch the graph.

  • Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
    • If the equation is in the form (x−h)2a2+(y−k)2b2=1\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1, where a>ba>b, then
      • the center is (h,k)\left(h,k\right)
      • the major axis is parallel to the x-axis
      • the coordinates of the vertices are (h±a,k)\left(h\pm a,k\right)
      • 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)
    • If the equation is in the form (x−h)2b2+(y−k)2a2=1\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}=1, where a>ba>b, then
      • the center is (h,k)\left(h,k\right)
      • the major axis is parallel to the y-axis
      • the coordinates of the vertices are (h,k±a)\left(h,k\pm a\right)
      • 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)
  • Solve for cc using the equation c2=a2−b2{c}^{2}={a}^{2}-{b}^{2}.
  • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

Example: Graphing an Ellipse Centered at (h, k)

Graph the ellipse given by the equation, (x+2)24+(y−5)29=1\dfrac{{\left(x+2\right)}^{2}}{4}+\dfrac{{\left(y - 5\right)}^{2}}{9}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: First, we determine the position of the major axis. Because 9>49>4, the major axis is parallel to the y-axis. Therefore, the equation is in the form (x−h)2b2+(y−k)2a2=1\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}=1, where b2=4{b}^{2}=4 and a2=9{a}^{2}=9. It follows that:

  • the center of the ellipse is (h,k)=(−2,5)\left(h,k\right)=\left(-2,\text{5}\right)
  • the coordinates of the vertices are (h,k±a)=(−2,5±9)=(−2,5±3)\left(h,k\pm a\right)=\left(-2,5\pm \sqrt{9}\right)=\left(-2,5\pm 3\right), or (−2,2)\left(-2,\text{2}\right) and (−2,8)\left(-2,\text{8}\right)
  • the coordinates of the co-vertices are (h±b,k)=(−2±4,5)=(−2±2,5)\left(h\pm b,k\right)=\left(-2\pm \sqrt{4},5\right)=\left(-2\pm 2,5\right), or (−4,5)\left(-4,5\right) and (0,5)\left(0,\text{5}\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}. Solving for cc, we have:

\begin{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{9 - 4} \\ &=\pm \sqrt{5} \end{align}

Therefore, the coordinates of the foci are (−2,5−5)\left(-2,\text{5}-\sqrt{5}\right) and (−2,5+5)\left(-2,\text{5+}\sqrt{5}\right). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation (x−4)236+(y−2)220=1\dfrac{{\left(x - 4\right)}^{2}}{36}+\dfrac{{\left(y - 2\right)}^{2}}{20}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: Center: (4,2)\left(4,2\right); vertices: (−2,2)\left(-2,2\right) and (10,2)\left(10,2\right); co-vertices: (4,2−25)\left(4,2 - 2\sqrt{5}\right) and (4,2+25)\left(4,2+2\sqrt{5}\right); foci: (0,2)\left(0,2\right) and (8,2)\left(8,2\right)

How To: Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form.

  • Recognize that an ellipse described by an equation in the form ax2+by2+cx+dy+e=0a{x}^{2}+b{y}^{2}+cx+dy+e=0 is in general form.
  • Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
  • Factor out the coefficients of the x2{x}^{2} and y2{y}^{2} terms in preparation for completing the square.
  • Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, m1(x−h)2+m2(y−k)2=m3{m}_{1}{\left(x-h\right)}^{2}+{m}_{2}{\left(y-k\right)}^{2}={m}_{3}, where m1,m2{m}_{1},{m}_{2}, and m3{m}_{3} are constants.
  • Divide both sides of the equation by the constant term to express the equation in standard form.

recall complete the square

To write the equation of an ellipse in standard form, you'll need to complete the square two times. Given an expression of the form a(x2+bx)a\left(x^2+bx\right), add (b2)2\left(\dfrac{b}{2}\right)^2 inside the parentheses, then subtract a(b2)2a\left(\dfrac{b}{2}\right)^2 to counteract the change you made. If completing the square on one side of an equation, you may either subtract the value of a(b2)2a\left(\dfrac{b}{2}\right)^2 from that side, or add it to the other to maintain equality. Then factor the perfect square trinomial you created inside the original parentheses. Example

a(x2+bx)\qquad a\left(x^2+bx\right)

=a(x2+bx+(b2)2)−a(b2)2=a\left(x^2+bx+ \left(\dfrac{b}{2}\right)^2\right)-a\left(\dfrac{b}{2}\right)^2

=a(x+b2)2−a(b2)2=a\left(x+ \dfrac{b}{2}\right)^2-a\left(\dfrac{b}{2}\right)^2

 

Example: Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form

Graph the ellipse given by the equation 4x2+9y2−40x+36y+100=04{x}^{2}+9{y}^{2}-40x+36y+100=0. Identify and label the center, vertices, co-vertices, and foci.

Answer: We must begin by rewriting the equation in standard form.

4x2+9y2−40x+36y+100=04{x}^{2}+9{y}^{2}-40x+36y+100=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation.

(4x2−40x)+(9y2+36y)=−100\left(4{x}^{2}-40x\right)+\left(9{y}^{2}+36y\right)=-100

Factor out the coefficients of the squared terms.

4(x2−10x)+9(y2+4y)=−1004\left({x}^{2}-10x\right)+9\left({y}^{2}+4y\right)=-100

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

4(x2−10x+25)+9(y2+4y+4)=−100+100+364\left({x}^{2}-10x+25\right)+9\left({y}^{2}+4y+4\right)=-100+100+36

Rewrite as perfect squares.

4(x−5)2+9(y+2)2=364{\left(x - 5\right)}^{2}+9{\left(y+2\right)}^{2}=36

Divide both sides by the constant term to place the equation in standard form.

(x−5)29+(y+2)24=1\dfrac{{\left(x - 5\right)}^{2}}{9}+\dfrac{{\left(y+2\right)}^{2}}{4}=1

Now that the equation is in standard form, we can determine the position of the major axis. Because 9>49>4, the major axis is parallel to the x-axis. Therefore, the equation is in the form (x−h)2a2+(y−k)2b2=1\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}+\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1, where a2=9{a}^{2}=9 and b2=4{b}^{2}=4. It follows that:
  • the center of the ellipse is (h,k)=(5,−2)\left(h,k\right)=\left(5,-2\right)
  • the coordinates of the vertices are (h±a,k)=(5±9,−2)=(5±3,−2)\left(h\pm a,k\right)=\left(5\pm \sqrt{9},-2\right)=\left(5\pm 3,-2\right), or (2,−2)\left(2,-2\right) and (8,−2)\left(8,-2\right)
  • the coordinates of the co-vertices are (h,k±b)=(5,−2±4)=(5,−2±2)\left(h,k\pm b\right)=\left(\text{5},-2\pm \sqrt{4}\right)=\left(\text{5},-2\pm 2\right), or (5,−4)\left(5,-4\right) and (5,0)\left(5,\text{0}\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}. Solving for cc, we have:

\begin{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{9 - 4} \\ &=\pm \sqrt{5} \end{align}

Therefore, the coordinates of the foci are (5−5,−2)\left(\text{5}-\sqrt{5},-2\right) and (5+5,−2)\left(\text{5+}\sqrt{5},-2\right).

Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse. 4x2+y2−24x+2y+21=04{x}^{2}+{y}^{2}-24x+2y+21=0

Answer: (x−3)24+(y+1)216=1[/latex];center:[latex](3,−1)[/latex];vertices:[latex](3,−5)[/latex]and[latex](3,3)[/latex];co−vertices:[latex](1,−1)[/latex]and[latex](5,−1)[/latex];foci:[latex](3,−1−23)[/latex]and[latex](3,−1+23)\dfrac{{\left(x - 3\right)}^{2}}{4}+\dfrac{{\left(y+1\right)}^{2}}{16}=1[/latex]; center: [latex]\left(3,-1\right)[/latex]; vertices: [latex]\left(3,-\text{5}\right)[/latex] and [latex]\left(3,\text{3}\right)[/latex]; co-vertices: [latex]\left(1,-1\right)[/latex] and [latex]\left(5,-1\right)[/latex]; foci: [latex]\left(3,-\text{1}-2\sqrt{3}\right)[/latex] and [latex]\left(3,-\text{1+}2\sqrt{3}\right)

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