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6.1.3: Ellipses Not Centered at the Origin

  • Page ID
    14749
  • Ellipses Centered at (h, k)

    Your homework assignment is to draw the ellipse \(\ 16(x-2)^{2}+4(y+3)^{2}=144\). What is the vertex of your graph and where will the foci of the ellipse be located?


    Ellipses Centered at (h,k)

    An ellipse does not always have to be placed with its center at the origin. If the center is \(\ (h, k)\) the entire ellipse will be shifted \(\ h\) units to the left or right and k units up or down. The equation becomes \(\ \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\). We will address how the vertices, co-vertices, and foci change in the following problem.

    Let's graph \(\ \frac{(x-3)^{2}}{16}+\frac{(y+1)^{2}}{4}=1\). Then, we'll find the vertices, co-vertices, and foci.

    First, we know this is a horizontal ellipse because \(\ 16>4\). Therefore, the center is \(\ (3,-1)\) and \(\ a=4\) and \(\ b=2\). Use this information to graph the ellipse.

    To graph, plot the center and then go out 4 units to the right and left and then up and down two units. This is also how you can find the vertices and co-vertices. The vertices are (3 ± 4, −1) or (7, −1) and (−1, −1). The co-vertices are (3, −1 ± 2) or (3, 1) and (3, −3).

    f-d_d9af0957c67bb6e0338c3f50c9bd7fc02b79dfc4cf846920ca92a85e+IMAGE_TINY+IMAGE_TINY.png

    To find the foci, we need to find \(\ c\) using \(\ c^{2}=a^{2}-b^{2}\). 

    \(\ \begin{aligned}
    c^{2} &=16-4=12 \\
    c &=2 \sqrt{3}
    \end{aligned}\)

    Therefore, the foci are \(\ (3 \pm 2 \sqrt{3},-1)\).

    From this problem, we can create formulas for finding the vertices, co-vertices, and foci of an ellipse with center \(\ (h, k)\). Also, when graphing an ellipse, not centered at the origin, make sure to plot the center.

    Orientation Equation Vertices Co-Vertices Foci
    Horizontal \(\ \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\) \(\ (h±a, k)\) \(\ (h, k±b)\) \(\ (h±c, k)\)
    Vertical \(\ \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1\) \(\ (h, k±a)\) \(\ (h±b, k)\) \(\ (h, k±c)\)

    Now, let's find the equation of the ellipse with vertices (−3, 2) and (7, 2) and co-vertex (2, −1).

    These two vertices create a horizontal major axis, making the ellipse horizontal. If you are unsure, plot the given information on a set of axes. To find the center, use the midpoint formula with the vertices.

    \(\ \left(\frac{-3+7}{2}, \frac{2+2}{2}\right)=\left(\frac{4}{2}, \frac{4}{2}\right)=(2,2)\)

    The distance from one of the vertices to the center is \(\ a,|7-2|=5\). The distance from the co-vertex to the center is \(\ b,|-1-2|=3\). Therefore, the equation is \(\ \frac{(x-2)^{2}}{5^{2}}+\frac{(y-2)^{2}}{3^{2}}=1\) or \(\ \frac{(x-2)^{2}}{25}+\frac{(y-2)^{2}}{9}=1\).

    Finally, let's graph \(\ 49(x-5)^{2}+25(y+2)^{2}=1225\) and find the foci.

    First we have to get this into standard form, like the equations above. To make the right side 1, we need to divide everything by 1225.

    \(\ \begin{array}{c}
    \frac{49(x-5)^{2}}{1225}+\frac{25(y+2)^{2}}{1225}=\frac{1225}{1225} \\
    \frac{(x-5)^{2}}{25}+\frac{(y+2)^{2}}{49}=1
    \end{array}\)

    Now, we know that the ellipse will be vertical because 25 < 49. a=7, b=5 and the center is (5,−2).

    f-d_55c0f3ed918f8b1f580abfdcd62e59e26d71e0a9b88f869aae0f46e9+IMAGE_TINY+IMAGE_TINY.png

    To find the foci, we first need to find \(\ c\) by using \(\ c^{2}=a^{2}-b^{2}\).

    \(\ \begin{array}{c}
    c^{2}&=49-25=24 \\
    c&=\sqrt{24}=2 \sqrt{6}
    \end{array}\)

    The foci are \(\ (5,-2 \pm 2 \sqrt{6})\) or \(\ (5,-6.9)\) and \(\ (5,2.9)\).


    Examples

    Example 1

    Earlier, you were asked to find the vertex of your graph and to determine where the foci of the ellipse will be located. 

    Solution

     

    We first need to get our equation in the form of \(\ \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\). So we divide both sides by 144.

    \(\ \begin{array}{r}
    \frac{16(x-2)^{2}}{144}+\frac{4(y+3)^{2}}{144}=\frac{144}{144} \\
    \frac{(x-2)^{2}}{9}+\frac{(y+3)^{2}}{36}
    \end{array}\)

    Now we can see that \(\ h=2\) and \(\ 3=-k\) or \(\ k=-3\). Therefore the origin is \(\ (2,-3)\).

    Because \(\ 9<36\), we know this is a vertical ellipse. To find the foci, use \(\ c^{2}=a^{2}-b^{2}\).

    \(\ \begin{aligned}
    c^{2} &=36-9=27 \\
    c &=\sqrt{27}=3 \sqrt{3}
    \end{aligned}\)

    The foci are therefore \(\ (2,-3+3 \sqrt{3})\) and \(\ (2,-3-3 \sqrt{3})\)

    Example 2

    Find the center, vertices, co-vertices and foci of \(\ \frac{(x+4)^{2}}{81}+\frac{(y-7)^{2}}{16}=1\).

    Solution

    The center is \(\ (-4,7), a=\sqrt{81}=9\) and \(\ b=\sqrt{16}=4\), making the ellipse horizontal. The vertices are \(\ (-4 \pm 9,7)\) or \(\ (-13,7)\) and \(\ (5,7)\). The co-vertices are \(\ (-4,7 \pm 4)\) or \(\ (−4,3)\) and \(\ (-4,11)\). Use \(\ c^{2}=a^{2}-b^{2}\) to find \(\ c\). 

    \(\ \begin{array}{l}
    c^{2}&=81-16=65 \\
    c&=\sqrt{65}
    \end{array}\)

    The foci are \(\ (-4-\sqrt{65}, 7)\) and \(\ (-4+\sqrt{65}, 7)\).

    Example 3

    Graph \(\ 25(x-3)^{2}+4(y-1)^{2}=100\) and find the foci.

    Solution

    Change this equation to standard form in order to graph.

    f-d_a85fc3a7c8dbaa8bed4d0eb990a54da2aba80bc7c73f51dc1b89112f+IMAGE_TINY+IMAGE_TINY.png

    \(\ \begin{array}{l}
    \frac{25(x-3)^{2}}{100}+\frac{4(y-1)^{2}}{100}=\frac{100}{100} \\
    \frac{(x-3)^{2}}{4}+\frac{(y-1)^{2}}{25}=1
    \end{array}\)

    center: \(\ (3,1), b=2, a=5\)

    Find the foci. 

    \(\ \begin{aligned}
    c^{2} &=25-4=21 \\
    c &=\sqrt{21}
    \end{aligned}\)

    The foci are \(\ (3,1+\sqrt{21})\) and \(\ (3,1-\sqrt{21})\).

    Example 4

    Find the equation of the ellipse with co-vertices (−3,−6) and (5,−6) and focus (1,−2).

    Solution

    The co-vertices (−3,−6) and (5,−6) are the endpoints of the minor axis. It is |−3−5|=8 units long, making b=4. The midpoint between the co-vertices is the center.

    \(\ \left(\frac{-3+5}{2},-6\right)=\left(\frac{2}{2},-6\right)=(1,-6)\)

    The focus is (1,−2) and the distance between it and the center is 4 units, or \(\ c\). Find \(\ a\).

    \(\ \begin{array}{l}
    16=a^{2}-16 \\
    32=a^{2} \\
    a=\sqrt{32}=4 \sqrt{2}
    \end{array}\)

    The equation of the ellipse is \(\ \frac{(x-1)^{2}}{16}+\frac{(y+6)^{2}}{32}=1\).


    Review

    Find the center, vertices, co-vertices, and foci of each ellipse below.

    1. \(\ \frac{(x+5)^{2}}{25}+\frac{(y+1)^{2}}{36}=1\)
    2. \(\ (x+2)^{2}+16(y-6)^{2}=16\)
    3. \(\ \frac{(x-2)^{2}}{9}+\frac{(y-3)^{2}}{49}=1\)
    4. \(\ 25 x^{2}+64(y-6)^{2}=1600\)
    5. \(\ (x-8)^{2}+\frac{(y-4)^{2}}{9}=1\)
    6. \(\ 81(x+4)^{2}+4(y+5)^{2}=324\)
    7. Graph the ellipse in #1.
    8. Graph the ellipse in #2.
    9. Graph the ellipse in #4.
    10. Graph the ellipse in #5.

    Using the information below, find the equation of each ellipse.

    1. vertices: (−2, −3) and (8, −3) co-vertex: (3, −5)
    2. vertices: (5, 6) and (5, −12) focus: (5, −7)
    3. co-vertices: (0, 4) and (14, 4) focus: (7, 1)
    4. foci: (−11, −4) and (1, −4) vertex: (−12, −4)
    5. Extension Rewrite the equation of the ellipse, 36x+ 25y− 72x + 200y − 464 = 0 in standard form, by completing the square for both the x and y terms.

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 10.6. 


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