SOLUTION: how do you convert this conic section into standard form? 4x^2+y^2-48x-4y+48= 0

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Question 702648: how do you convert this conic section into standard form? 4x^2+y^2-48x-4y+48= 0
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
Here's one like it you can use as a model to go by:

4x² + 9y² - 64x + 162y + 949 = 0

We want to get it in one of these forms:

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1 or %28x-h%29%5E2%2Fb%5E2%2B%28y-k%29%5E2%2Fa%5E2=1 

4x² + 9y² - 64x + 162y + 949 = 0

Swap the second and third terms to get the y-terms 
together and the x terms together.  Also add -949 
to both sides:

      4x² - 64x + 9y² + 162y = -949 

Factor just the 4 out of the first two terms on the 
left, and just the 9 out of the last two terms on
the left:

   4(x² - 16x) + 9(y² + 18y) = -949

We find what numbers we must add inside
those parentheses to complete the square.

Multiply the coefficient of x in the first parentheses,
which is -16 by1%2F2.  Get -8.  The square that and
get (-8)² = +64.  That's what we add inside the first 
parentheses.

Multiply the coefficient of y in the second parentheses,
which is 18 by1%2F2.  Get 9.  The square that and
get 9² = +81.  That's what we add inside the second 
parentheses.

   4(x² - 16x + 64) + 9(y² + 18y + 81) = -949 + 256 + 729

Notice that we added 256 on the right because when +64 
is added inside the first parentheses, since there is a 4
coefficient before the entire first parentheses, it really 
amounts to adding 4×64 or 256 to the left side, so we have 
to do the same to the right side.

Notice also that we added 729 on the right because when +81 
is added inside the second parentheses, since there is a 9
coefficient before the entire first parentheses, it really 
amounts to adding 9×81 or 729 to the left side, so we have 
to do the same to the right side.

Now we factor the two parentheses and combine the terms on
the right side:

     4(x - 8)(x - 8) + 9(y + 9)(y + 9) = 36

                 4(x - 8)² + 9(y + 9)² = 36

Divide through by 36 to get 1 on the right:

                 %28x+-+8%29%5E2%2F36 + 9%28y+%2B+9%29%5E2%2F36 = 36%2F36
               
                 %28x+-+8%29%5E2%2F36 + %28y+%2B+9%29%5E2%2F4 = 1

That's the form  %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1

where (h,k) = (8,-9), a = 6, b = 2 and so the graph is

        

Edwin




Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
4x%5E2%2By%5E2-48x-4y%2B48=+0....reorder terms

4x%5E2-48x%2By%5E2-4y%2B48=+0...replace 48 with 144%2B4-100
4x%5E2-48x%2By%5E2-4y%2B144%2B4-100=+0.......group
%284x%5E2-48x%2B144%29%2B%28y%5E2-4y%2B4%29-100=0

4%28x%5E2-12x%2B36%29%2B%28y%5E2-4x%2B4%29-100=0....write %28x%5E2-12x%2B36%29 as %28x-6%29%5E2 and
%28y%5E2-4x%2B4%29 as %28y-2%29%5E2

4%28x-6%29%5E2%2B%28y-2%29%5E2-100=0

4%28x-6%29%5E2%2B%28y-2%29%5E2=100

4%28x-6%29%5E2%2F100%2B%28y-2%29%5E2%2F100=100%2F100

cross%284%29%28x-6%29%5E2%2Fcross%28100%2925%2B%28y-2%29%5E2%2F100=1

%28x-6%29%5E2%2F25%2B%28y-2%29%5E2%2F100=1

%28x-6%29%5E2%2F5%5E2%2B%28y-2%29%5E2%2F10%5E2=1....this is your ellipse with a center (h,k at (6,2, semi-minor axis length a=5, and semi-major axis length b=10