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Question 663432: My teacher put this problem and solution on the board and I need a laymans termed explaination of why he did certain things in the solution.
3x^2+4y^2-6x-24y+39=0
3(x^2-2x+1)+4(y^2-6y+1)=0-39+3+36
3(x-1)^2+4(y-3)^2=0
So, I get confused when my teacher added 1 inside both of the parenthesis...why did he do that? Where did he get the 1 from? Why not just leave the 1 out of both of the parenthesis? Please help!
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! My teacher put this problem and solution on the board and I need a laymans termed explaination of why he did certain things in the solution.
3x^2+4y^2-6x-24y+39=0
3(x^2-2x+1)+4(y^2-6y+1)=0-39+3+36
3(x-1)^2+4(y-3)^2=0
So, I get confused when my teacher added 1 inside both of the parenthesis...why did he do that? Where did he get the 1 from? Why not just leave the 1 out of both of the parenthesis?
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3x^2+4y^2-6x-24y+39=0
He completed the squares of the x & y terms.
3x^2+4y^2-6x-24y+39=0
3x^2 - 6x + 4y^2 - 24y = -39
3(x^2 + 2x) + 4(y^2 - 6y) = -39
3(x^2 + 2x + 1) + 4(y^2 - 6y + 9) = -39 + 3*1 + 4*9 = 0
3(x+1)^2 + 4(y-3)^2 = 0
Divide by 3*4 = 12
 (x+1)^2/4 + (y-3)^2/3 = 0
That's the standard form of an ellipse, but the right side = 0, so it's a "point ellipse."
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Not a good example, imo.
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PS A lot of teachers don't have a good grasp of math.
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3(x^2-2x+1)+4(y^2-6y+1)=0-39+3+36
3(x-1)^2+4(y-3)^2=0
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
He didn't add 1 inside of both sets of parentheses, unless he just made a mistake that is. However, the 36 term added to the RHS tells me otherwise.
What he is doing is completing the square on each of the variables. The goal, when you have a general form equation of an ellipse like this is to convert it to the standard form, namely: ^2}{a^2}\ +\ \frac{(y\ -\ k)^2}{b^2}\ =\ 1) which is a ellipse centered at ) with semi-major and semi-minor axes  and  (the major axis is larger).
To perform this conversion, we have to complete the square.
Step 1: Move the constant term into the RHS:
Step 2: Put the  terms together and the  terms together
Step 3: Factor the coefficient on the  term out of the terms, leaving a spot inside the parentheses, then do the same thing for the terms:
\ +\ 4\left(y^2\ -\ 6y\ +\ ??\right)\ =\ -\ 39)
Step 4: Complete the square on . Divide the coefficient on by 2, and square the result. -2 divided by 2 is -1, -1 squared is 1, and therefore 1 is the number that needs to be added inside of the parentheses. Note that the 3 coefficient on the outside of the parentheses means that you are actually adding 3 to the LHS when you insert a 1 into the set of parentheses containing the term. Hence you have to compensate by adding 3 to the RHS.
\ +\ 4\left(y^2\ -\ 6y\ +\ ??\right)\ =\ -\ 39\ +\ 3)
Step 5: Do the part. -6 divided by 2 is -3. -3 squared is 9. 9 inside the parentheses and 9 times 4 = 36 in the RHS:
\ +\ 4\left(y^2\ -\ 6y\ +\ 9\right)\ =\ -\ 39\ +\ 3\ +\ 36)
Step 6:
Factor the expressions in the parentheticals, which should be easy to do because we arranged for them to be perfect squares. Also collect terms in the RHS.
^2\ +\ 4\left(y\ -\ 3\right)^2\ =\ 0)
Step 7:
Gaack! The RHS is zero! We won't have an ellipse, we will have a point. Be that as it may, your RHS might not be zero next time. Proceed:
Ordinarily you would divide by the simplified RHS to produce a 1 in the RHS, but that is not to be in this case. Divide by the LCM of the coefficients on the parenthetical expressions:
^2}{4}\ +\ \frac{\left(y\ -\ 3\right)^2}{3}\ =\ 0)
And enjoy your graph of a point. By the way: Super Double WoWWie Extra Credit. What are the coordinates of that point?
Step 8: Don't doze off in class any more.
John
My calculator said it, I believe it, that settles it
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