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Math Help - Ellipses

  1. #1
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    Ellipses

    Can you please check my answers?

    Graph the ellipse and locate the foci.

    9x^2=144-16y^2

    I got


    x^2/5+y^2/9=1
    I got



    Find the standard form of the equation of the ellipse and give the location of its foci.



    I got: x^2/9 +y^2/4=1
    foci at: (- squareroot of 5,0) and (squareroot 5,0)


    I got: x^2/36+y^2/4=1
    foci at: (0,-4 squareroot 2) and (0, 4 squareroot 2)
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  2. #2
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    Hello, soly_sol!

    Can you please check my answers?

    Graph the ellipse and locate the foci: . 9x^2\:=\:144-16y^2

    I got
    . . . . no

    We have: . 9x^2 + 16y^2 \:=\:144

    Divide by 144: . \frac{x^2}{16} + \frac{y^2}{9} \;=\;1

    Hence: . a = 4,\;b = 3

    The ellipse is 8 units wide and 6 units high.



    \frac{x^2}{5} + \frac{y^2}{9} \;=\;1

    I got
    . . . . Good!

    Find the standard form of the equation of the ellipse and give the location of its foci.



    I got: . \frac{x^2}{9} + \frac{y^2}{4} \;=\;1 . . . . Yes!

    Foci at: . (\pm\sqrt{5},\:0) . . . . Right!



    I got: . \frac{x^2}{36} + \frac{y^2}{4} \;=\;1 . . . . no

    Foci at: . (0,\:\pm4\sqrt{2})
    The ellipse is 4 units wide and 12 units high.
    We have: . a = 2,\;b = 6

    The equation is: . \frac{x^2}{4} + \frac{y^2}{36} \;=\;1



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  3. #3
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    x^2/36+y^2/4=1

    On this problem, I dont understand why it would be x^2/4 +y^2/36=1 ?
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  4. #4
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    I dont understand how to do this problem. Can you please help me?

    Find the standard form of the equation of the ellipse and give the location of its foci.



    center at (-1, 3)
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by soly_sol View Post
    I dont understand how to do this problem. Can you please help me?

    Find the standard form of the equation of the ellipse and give the location of its foci.



    center at (-1, 3)
    Can you find the sizes of the semi-major and semi-minor axes from this diagram?

    -Dan
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  6. #6
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    Would the semi-major be (3,-3) and semi-minor be (-1,-6)?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by soly_sol View Post
    Would the semi-major be (3,-3) and semi-minor be (-1,-6)?
    Huh? The semi-major and minor axes are distances. Just like the radius of a circle. The semi-major axis (a) is half of the "long axis" of the ellipse and the semi-minor axis (b) is half of the "short axis." So I'm getting a = 4 and along the x direction and b = 3 along the y direction. So the form for your ellipse will be:
    \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
    for this form of ellipse, where (h, k) is the center point of the ellipse.

    -Dan
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  8. #8
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    Can you please explain to me how to get the semi-major and semi-minor?

    Also, this is the answer I got:

    ((x+1)^2)/(16) + ((y+3)^2)/(9)
    foci: (-1 + squareroot 7, -3) and (-1 negative squareroot 7,-3)
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by soly_sol View Post
    Can you please explain to me how to get the semi-major and semi-minor?
    If you are having this many problems about how to do this you really need to speak with your teacher for extra help.

    I simply looked at the ellipse, and noted it was wider than it was tall. So the horizontal axis is the major axis. Then I noted the ellipse is 8 units across, so the semi-major axis has a length of 4. Similarly for the vertical axis.

    -Dan
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