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Homework Help: Identify Type of Graph

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    4x2 - 8x + 3y2 + 18y - 23 = 0

    2. Relevant equations

    3. The attempt at a solution


    What did I do wrong?
  2. jcsd
  3. Apr 17, 2010 #2
    Your factors are incorrect. Your graph is not a circle.
  4. Apr 17, 2010 #3
    Can I get assistance to break it down from a to z?
  5. Apr 17, 2010 #4
    Well, sure. We can both learn as we go since I'm only familiar with parabolas and circles. Let's both work on our ends to find the solution.

    To start, re-check your equation and make sure you typed it as it appears on your notebook or textbook, please. In the meantime, we are looking for something of the form:

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


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

    That should look very familiar to the circle equation, no?
  6. Apr 18, 2010 #5


    Staff: Mentor

    This is NOT the equation of a circle, if that's what you were driving at. The equation you gave is similar to (not familiar to) the equation of a circle, but the equation is of an ellipse.
    4x2 - 8x + 3y2 + 18y - 23 = 0

    You started off on the right track, but you complete the square in the x and y terms incorrectly.

    4x2 - 8x + 3y2 + 18y - 23 = 0
    4(x2 - 2x) + 3(y2 + 6y) = 23
    Now figure out what you need to add in the first group to complete the square in the x terms, and what you need to add in the second group to complete the square in the y terms. Be sure to add both numbers to the right side.

    When you have done that, divide both sides by whatever number you have on the right side. You should end up with something that looks like this:
    [tex]\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1[/tex]
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