SOLUTION: find the coordinates of the center of the elipse and the values of a and b. 2x^2 + 3y^2 - 6x + 6y = 12

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Question 637272: find the coordinates of the center of the elipse and the values of a and b.
2x^2 + 3y^2 - 6x + 6y = 12
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2%2B3y%5E2-6x%2B6y=12
  1. We start by completing the squares:
    1. Moving the constant term to the other side of the equation. But your constant term is already on the other side.
    2. Rearrange the terms so that the x's and y's are together:
      2x%5E2-6x%2B3y%5E2%2B6y=12
    3. Factor out the squared-term coefficients:
      2%28x%5E2-3x%29%2B3%28y%5E2%2B2y%29=12
    4. Find 1/2 of the coefficients of the 1st-power terms.
      For the x-term, half of -3 is -3/2
      For the y-term, half of 2 is 1
      Remember these for later.
    5. Square each of these halves:
      %28-3%2F2%29%5E2+=+9%2F4
      %281%29%5E2+=+1
    6. Add to both sides so that the squares from the last step can be added in the parentheses. This is usually the trickiest step. Because of the 2 in front of 2%28x%5E2-3x%29 we need to add 2 (9/4)'s to each side in order for us to be able to write 2%28x%5E2-3x%2B9%2F4%29 And because of the 3 in front of 3%28y%5E2%2B2y%29 we will need to add 3 1's to each side in order for us to write 3%28y%5E2%2B2y%2B1%29. All together we get:
      2%28x%5E2-3x%2B9%2F4%29%2B3%28y%5E2%2B2y%2B1%29=12%2B2%2A%289%2F4%29%2B3%2A1
      The right side simplifies:
      2%28x%5E2-3x%2B9%2F4%29%2B3%28y%5E2%2B2y%2B1%29=12%2B9%2F2%29%2B3
      2%28x%5E2-3x%2B9%2F4%29%2B3%28y%5E2%2B2y%2B1%29=24%2F2%2B9%2F2%2B6%2F2
      2%28x%5E2-3x%2B9%2F4%29%2B3%28y%5E2%2B2y%2B1%29=39%2F2
      And we can eliminate the fraction by multiplying both sides by 2:
      4%28x%5E2-3x%2B9%2F4%29%2B6%28y%5E2%2B2y%2B1%29=39
    7. At this point we have completed the squares. All that is left is to rewrite the trinomials as a square. For this we write (x [or y] plus the "half" from step 4)^2:
      4%28x%2B%28-3%2F2%29%29%2B6%28y%2B1%29%5E2=39
      which simplifies to:
      4%28x-3%2F2%29%5E2%2B6%28y%2B1%29%5E2=39
  2. Make the right side a 1. This is done by dividing both sides by the number we see there, 39:
    4%28x-3%2F2%29%5E2%2F39%2B6%28y%2B1%29%5E2%2F39=1
  3. Next, we need to eliminate any coefficients in front of the squares in the numerators. We do this by multiplying the numerator and denominator by the reciprocal of the coefficient:

    which simplifies to:
    %28x-3%2F2%29%5E2%2F%2839%2F4%29%2B%28y%2B1%29%5E2%2F%2839%2F6%29=1
  4. Last of all, if any of the squared binomials are additions, change them to an equivalent subtraction:
    %28x-3%2F2%29%5E2%2F%2839%2F4%29%2B%28y-%28-1%29%29%5E2%2F%2839%2F6%29=1

As weird as
%28x-3%2F2%29%5E2%2F%2839%2F4%29%2B%28y-%28-1%29%29%5E2%2F%2839%2F6%29=1
may look this is the form of the equation that tells us what we are looking for. In the numerators we can "read" the coordinates of the center:
(3/2, -1)

And we can get the "a" and "b" from the denominators. The denominators are a%5E2 and b%5E2. We must figure out which is which. In an ellipse, a > b which also means a%5E2+%3E+b%5E2. Looking at our denominators we should be able to see that 39/4 is larger than 39/6. (Remember, smaller denominators make bigger fractions.)

So 39/4 is a%5E2 making...
a+=+sqrt%2839%2F4%29+=+sqrt%2839%29%2Fsqrt%284%29+=+sqrt%2839%29%2F2
And 39/6 is b%5E2 making...