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Math Help - ellipse enigmas! - urgent

  1. #1
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    Post ellipse enigmas! - urgent

    Hey, I got a few ellipse problems I'm havin trouble with. Any help appreciated.

    Write in standard form, showing work. (standard form is (x^2 / a^2) + (y^2 / b^2) = 1, right?)
    1. 16x^2 + 25y^2 - 64x - 100y + 564 = 0

    2. 5x^2 + 3y^2 - 40x - 36y + 188 = 0

    3. 3x^2 + 4y^2 - 12x + 16y + 16 = 0

    Find the equation of an ellipse that has foci (0, +-2) and the endpoints of the minor axes are (+-5,0). Write in standard form, showing work.
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  2. #2
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    Quote Originally Posted by leviathanwave View Post
    1. 16x^2 + 25y^2 - 64x - 100y + 564 = 0
    16(x^2-4x)+25(y^4-4y)=-564

    16(x^2-4x+4)+25(y^4-4y+4)=-564+64+100=-400

    16(x-2)^2+25(y-2)^2=-400

    This is nothing.
    Check your problem.

    I think #2 is the same situation.
    Now wonder they are enigmas.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Hey, I got a few ellipse problems I'm havin trouble with. Any help appreciated.

    Write in standard form, showing work. (standard form is (x^2 / a^2) + (y^2 / b^2) = 1, right?)
    the standard form of an ellipse with center (h,k) is:

    [(x - h)^2]/a^2 + [(y - k)^2]/b^2 = 1

    the formula you wrote is the standard form of an ellipse with center the origin.

    starting from the equations you have given, we can get to the standard form by completing the square as you saw TPH did. however, he ended up with a negative number for the constant, so should he try to get 1 on one side, he would have to multiply by a negative number and the result wouldn't be the equation of an ellipse. check your questions again.
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  4. #4
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    Post hmmm....???

    That's the thing, I double checked em before i put em on, now I'm lookin right at them, and that's what they are... what in the world??
    well... any ideas on the "find the equation..." or the third one?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Find the equation of an ellipse that has foci (0, +-2) and the endpoints of the minor axes are (+-5,0). Write in standard form, showing work.
    we see that the minor axis lies along the x-axis. the center is halfway between the endpoints of the axis. so here, the center is the origin (0,0).

    the equation of an ellipse with center (0,0), and foci (0, +/- c) is:

    x^2/b^2 + y^2/a^2 = 1

    the minor axis is 2b = 10 (since the minor axis is 10 units long)
    => b = 5

    the foci are (0, +/- 2), where c = +/-2.
    now c^2 = a^2 - b^2
    => 4 = a^2 - 25
    => a^2 = 29

    so our ellipse is:

    x^2/25 + y^2/29 = 1
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    since we're apparently having problems with the first, let's try to find the equation for the other 2.

    Quote Originally Posted by leviathanwave View Post
    2. 5x^2 + 3y^2 - 40x - 36y + 188 = 0
    5x^2 + 3y^2 - 40x - 36y + 188 = 0

    we proceed by completing the square:

    5x^2 - 40x + 3y^2 - 36y = -188 ...........grouped terms with x's and y's togaether

    => 5(x^2 - 8x) + 3(y^2 - 12y) = -188
    => 5(x^2 - 8x + (-4)^2 - (-4)^2) + 3(y^2 - 12y + (-6)^2 - (-6)^2) = -188
    => 5[(x - 4)^2 - 16] + 3[(y - 6)^2 - 36] = -188
    => 5(x - 4)^2 - 80 + 3(y - 6)^2 - 108 = -188
    => 5(x - 4)^2 + 3(y - 6)^2 = 0

    so we seem to be having a problem for this one as well.


    3. 3x^2 + 4y^2 - 12x + 16y + 16 = 0
    3x^2 - 12x + 4y^2 + 16y + 16 = 0
    => 3(x^2 - 4x) + 4(y^2 + 4y) + 16 = 0
    => 3(x^2 - 4x + (-2)^2 - (-2)^2) + 4(y^2 + 4y + (2)^2 - (2)^2) + 16 = 0
    => 3[(x - 2)^2 - 4] + 4[(y + 2)^2 - 4] + 16 = 0
    => 3(x - 2)^2 - 12 + 4(y + 2)^2 - 16 + 16 = 0
    => 3(x - 2)^2 + 4(y + 2)^2 = 12
    => [(x - 2)^2]/4 + [(y + 2)^2]/3 = 1

    this one seems to work out ok.


    I'll recheck my work for the first and second questions and see if i made a mistake
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