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-   -   Conics, ellipses and hyperbolas (https://www.askmehelpdesk.com/showthread.php?t=209341)

  • Apr 24, 2008, 09:14 PM
    herz
    Conics, ellipses and hyperbolas
    I've got the circles and parabolas but I am having trouble with these... any guidance would be greatly appreciated.

    Ellipse: x^2 + 4y^2 + 6x + 16y + 21 = 0
    (x^2 + 6x +9) + 4y^2+16y) = -21 +9
    (x+3)^2 + 4(y^2 + 4y+4) = -12 + 16
    (x+3)^2 + 4(y+2)^2=4
    divide both sides by 4
    (x+3)^2 / 2^2 + (y+2)^2 / 1^2= 1
    Center (-3,-2)
    What next? Is the above correct?

    Hyperbola: 4y^2-9x^2+8y-54x-81=0
    (4y^2+8y ) - (9x^2-54x ) = 81
    4(y^2+2y+1) - 9(x^2-6x+9) = 81+4-81
    4(y+1)^2 - 9(x-3)^2=4
    divide both sides by 4
    (y+1)^2 / 1^2 - 9(x-3)^2 / 2^2 = 1
    here I wasn't sure what to do with the 9
  • Apr 28, 2008, 11:59 AM
    iamthetman
    Your ellipse looks perfect. The next step might be to graph it.

    Your hyperbola has a mistake in it. Going from line 1 to line 2 you forgot to change -54x to +54x since you factored out the negative sign outside the brackets.
  • Apr 29, 2008, 08:53 AM
    herz
    Yes but how do I get rid of the 9 on the second side? I ended up with 9(x+3)^2 / 4 I'm not sure how to handle that?
  • Apr 29, 2008, 09:39 AM
    ebaines
    You don't need to get rid of the 9. The standard form you want to put this in is like this:



    where (a,b,) is the center of the hyperbola, is the semi-major axis and is the semi-minor axis. If you combine the 9 from the denominator and the 4 in the numerator into a single denominator you have 4/9, or (2/3)^2. So 2/3 is the semi-major axis of the hyperbola, and 1 is the semi-minor axis.

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