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?
---------------------
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."
----------------------------
Not a good example, imo.
===========================
PS  A lot of teachers don't have a good grasp of math.
===========================
3(x^2-2x+1)+4(y^2-6y+1)=0-39+3+36
3(x-1)^2+4(y-3)^2=0