How do you complete the square with an ellipse equation?

Question

Example: x(squared)+2y(squared)-2x+8y-11=0
I am homeschooled and cannot figure out this completeing the square for nothing. The book doesn't explain how it comes up with its numbers. Thanks if you can help.

Answer 1

First you must re-group your terms (x together, y together, and the independent number on the other side of the equal sign)... this is:

x^2 - 2x +2y^2 + 8y = 1
then you factor a 2, from the y group:
x^2 - 2x +2(y^2 + 4y ) = 1
now look at the terms with the "not squared variables), take the coeficient (2 in the x case, 4 in the y case), divide them by two (which gives you 1 in the x, 2 in the y), and square those numbers (giving 1 in the x, 4 in the y), and now you put them--by adding them-- in the left side (where the spaces were left)... remembering to balance the equation... that is, for the x term (that we have a 1)
(x^2 - 2x + 1) +2(y^2 + 4y ) = 1 + 1
and for the y group, (that we had a 4... but when we put it inside the left parenthesis, since it is being multiplied by 2, you are REALLY adding an 8 in the left, which you have to add to the right to balance...
(x^2 - 2x + 1) +2(y^2+4y+4) = 1 + 1+ 8
and now you apply your perfect square factoring rule:
(x - 1)^2 + 2(y + 2)^2 = 10

and now to get the ellipse simplified form, you divide everything by 10....
(x-1)^2 / 10 + (y+2)^2 / 5 = 1
a horizontal ellipse with center in (1, -2)

Answer 2

I'll assume that the right side of the equation is zero.

Let's group terms.

x^2-2x-1 + 2y^2+8y = 0

The first set.: Looks like our work has already been done for us, that's already nearly a perfect square, if the sign on the one were switched, we'd have a perfect square. We can accomplish this by adding two and subtracting two (an operation which is the same as adding zero):

x^2-2x-1 + 2 - 2 Now combine the -1+2
x^2-2x+1 - 2 = (x-1)^2 - 2

Let's take the two to the other side so we don't have to worry about it for now

Beautiful, what do we have now?

(x-1)^2 + 2y^2+8y = 2 (Remember we took the -2 to the other side)

Now, let's make life a little easier on us and let's divide by two to make the coefficent of y^2 one. (That's a good trick to use when factoring)

[(x-1)^2]/'2 + y^2+4y = 1

Okay, so what do we do with these terms. Complete the square. How in the hell do we do that? Okay, take a general term like this:

y^2+by if we add b^2/4 and subtract it (we can do this because the total operation adds up to zero) we get:

y^2 + by + b^2/4 - b^2/4

Let's look at the first three terms, if we suggest a factoring of

(y+b/2)^2 = y^2 + by + b^2/4

Hey it works! Let's apply it to what we have:

y^2+4y+4 - 4
(y+2)^2 - 4

Notice that it leaves us with a - 4 to deal with. So what do we have:

[(x-1)^2]/2+(y+2) - 4 = 1

Standard form would suggest that we take the constant to the other side and then divide both sides of it so the right side is equal to one. Doing this, we have:

[(x-1)^2]/10+[(y+2)^2]/5 = 1

Which gives us the equation of an ellipse with a = 10, b = 5, and centered at (1,-2)

If you'd like to learn a little more about it:
http://mathworld.wolfram.com/Ellipse.html

Answer 3

x^2 + 2y^2 - 2x + 8y - 1 = 0
x^2 - 2x + 2y^2 + 8y - 1 = 0
(x^2 - 2x) + (2y^2 + 8y) - 1 = 0
(x^2 - 2x + 1 - 1) + 2(y^2 + 4y) - 1 = 0
((x - 1)^2 - 1) + 2(y^2 + 4y + 4 - 4) - 1 = 0
(x - 1)^2 - 1 + 2((y + 2)^2 - 4) - 1 = 0
(x - 1)^2 + 2(y + 2)^2 - 8 - 2 = 0
(x - 1)^2 + 2(y + 2)^2 - 10 = 0
(x - 1)^2 + 2(y + 2)^2 = 10
(((x - 1)^2)/10) + (((y + 2)^2)/5) = 1

in

(x^2 - 2x + 1 - 1) + 2(y^2 + 4y) - 1 = 0
((x - 1)^2 - 1) + 2(y^2 + 4y + 4 - 4) - 1 = 0

what i did was found half of 2, and squared it, and when you add it to both sides, it will be negative on the left side, but positive on the right side. Same goes for the 4 in y^2 + 4y