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Math Help - Graphing Off-Center Conics

  1. #1
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    Graphing Off-Center Conics

    I'm a bit lost at the moment

    I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but the teacher refused to teach me the subject before or after class. I'm going in tomorrow before school to talk with her, but in the meantime, I could use some help with this.

    For instance;

    How do I write an equation and graph the ellipse with foci at (4,2) and vertices at (4,4) and (4,-8) ?

    I understand standard form, but I don't understand how to add the center points.

    Thanks in advance to everyone!
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  2. #2
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    Hello ScottStedman

    Welcome to Math Help Forum!
    Quote Originally Posted by ScottStedman View Post
    I'm a bit lost at the moment

    I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but the teacher refused to teach me the subject before or after class. I'm going in tomorrow before school to talk with her, but in the meantime, I could use some help with this.

    For instance;

    How do I write an equation and graph the ellipse with foci at (4,2) and vertices at (4,4) and (4,-8) ?

    I understand standard form, but I don't understand how to add the center points.

    Thanks in advance to everyone!
    If the centre of the ellipse is at (h, k) then the equation is:

    \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

    where 2a and 2b are, as usual, the lengths of the axes. This is simply because (x-h) = 0 when x =h and (y-k) = 0 when y = k, so the centre has shifted from (0,0) to (h,k).

    In the example you've given, the centre is at the mid-point of the line joining (4,4) to (4,-8); i.e. at (4,-2). So h = 4, k =-2.

    The length of the vertical axis is 12 (the distance between these vertices), so b=6.

    The foci always lie on the major axis, so since one focus is at (4,2), the major axis is vertical, and the foci are at a distance be from the centre. So be = 4. And, when the major axis is vertical, a^2 = b^2-b^2e^2 = 36 - 16=20.

    Putting all this together, I reckon the equation is:

    \frac{(x-4)^2}{20} + \frac{(y+2)^2}{36} = 1

    I hope that answers some of your questions.

    Grandad
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  3. #3
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    Quote Originally Posted by ScottStedman View Post
    I was unfortunately absent on the day of graphing off-center conics in our Algebra 2 class. I've got the notes, but...I could use some help with this.
    To learn, in general, how to find an ellipse's equation from given information, try some online lessons.
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello ScottStedman

    I hope that answers some of your questions.

    Grandad
    Nice to see a forum where people actually care and are willing to help

    This was tremendously helpful. Thank you very much.
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