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Math Help - Equation of an Ellipse

  1. #1
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    Question Equation of an Ellipse

    Find an equation of an ellipse satisfying the given conditions.

    Foci: (-1,6) and (-1,0); Length of major axis: 10

    Need more of a check on my answer. If it's wrong, then I'll need some explanations on how to do the problem. If right, then that's all I really need.

    The center is obviously the midpoint between the 2 Foci, so it's (-1,3) and it's vertical.

    So it's equation is:

    (x+1)^2/b^2 + (y-3)^2/a^2 = 1

    A is half the length of the major axis (10) so a would = 5.
    A squared = 25

    C is x in the foci, so it's 1.
    C squared = 1

    So to find b^2, I need to plug values into:

    C^2 = a^2 - b^2
    1=25-b^2
    b^2 = 24

    So, assuming all of the above is correct, the final equation should be:

    (x+1)^2/24 + (y-3)^2/25 = 1

    Correct? Thanks.
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  2. #2
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    Hello, IAmLegendToo!

    Find an equation of an ellipse satisfying the given conditions.

    Foci: (-1,6) and (-1,0); Length of major axis: 10

    Need more of a check on my answer.

    The center is obviously the midpoint between the 2 Foci, so it's (-1,3) and it's vertical.

    So it's equation is: . \frac{(x+1)^2}{b^2} + \frac{(y-3)^2}{a^2} \;=\; 1

    a is half the length of the major axis (10) so: . a = 5 \quad\Rightarrow\quad a^2 = 25 . all correct!

    C is x in the foci, so it's 1 . . . . no

    c is the distance from the center to a focus: . c = 3

    Give it another try . . .

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  3. #3
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    Ahh, that sounds more correct. I don't know why I thought of it as simply X in a foci. Hmm.

    So then it's:

    C=3
    C^2 = 9
    9 = 25 - b^2
    b^2 = 16
    b= 4

    So, then equation would be:

    (x+1)^2/16 + (y-3)^2/25 = 1

    Correct now?
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