Question 474815: I need a brush-up on completing the square. I am working with equations for ellipses. I just need help getting this problem into the right form for an ellipse by completing the square.
Here is an example:
x^2-2x+4y^2+8y+1=0
I know that eventually this ends up looking like (x^2-2x+1)+4(y^2+2y+1)=-1+1+4(1). I do not understand how we get here. I know that we are supposes to multiply both(?) sides by the square of half of the x coefficients, but I'm still confused. Anyway, an explanation for a dummy like me would be super awesome. Thanks!
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
x²-2x+4y²+8y+1 = 0
The object is to make it look like this:
(x-h)² (y-k)²
—————— + ——————— = 1
a² b²
which is an ellipse that looks like this "ᆼ"
like an egg on a table, or:
(x-h)² (y-k)²
—————— + ——————— = 1
b² a²
which is an ellipse that looks like the number "0".
We can tell because a² > b²
We start with this:
x²-2x+4y²+8y+1 = 0
We get the 1 on the other side as a -1:
x²-2x+4y²+8y = -1
It's good that the terms are in the order they are.
Sometimes we have to switch them so that the terms
are in the order "x², x, y², y" as they are here.
Write it this way:
[x²-2x]+[4y²+8y] = -1
The coefficient of x² is 1 so we don't do anything yet to
the first bracket.
But the coefficient of y² is 4, so let's factor that out
in the 2nd bracket.
[x²-2x]+[4(y²+2y)] = -1
Now I'll dispense with the brackets and just have parentheses:
(x²-2x)+4(y²+2y) = -1
Next we want to make those two binomials into trinomials.
We skip some space after those binomials
(x²-2x )+4(y²+2y ) = -1
so we can add a number in those two boxes to make those
binomials into trinomials so they'll factor into squares
of binomials.
(x²-2x+)+4(y²+2y+) = -1
Now let's figure out what number goes in the first box.
The coefficient of x is -2 so we take half of it, getting -1,
then we square -1, getting (-1)² or 1,
so we add 1 where the first box is, but we also have to add
1 to the other side of the equation, like this:
[x²-2x+4]+4(y²+2y+) = -1+1
Now let's figure out what number goes in the second box.
The coefficient of y is 2 so we take half of it, getting 1,
then we square 1, getting 1² or 1, but wait! See the 4 in
front of the second parentheses? If we put a 1 in that box,
It will get multiplied by the 4 in front of the parentheses.
In other words putting a 1 in that second box will in effect
amount to the same as adding 4 times 1 or 4 to the left side,
not just 1. So we have to add 4(1) to the right side to offset
adding 1 inside that parentheses since it will be multiplied
by the 4, so we have:
(x²-2x+4)+4(y²+2y+1) = -1+1+4(1)
Notice that what's in the first parentheses,
x²-2x+4 factors as (x-2)(x-2) or (x-2)²
Also notice that what's in the second parentheses
y²+2y+1 factors as (y+1)(y+1) or (y+1)².
So this
(x²-2x+4)+4(y²+2y+1) = -1+1+4(1)
after substituting their factorization for the parentheses
and combining the terms on the right, we have
(x-2)²+4(y+1)² = 4
Next we get a 1 on the right by dividing all three terms by 4:
(x-2)² 4(y+1)² 4
—————— + ——————— = ———
4 4 4
And that simplifies to:
(x-2)² (y+1)²
—————— + ——————— = 1
4 1
which is in the form:
(x-h)² (y-k)²
—————— + ——————— = 1
a² b²
because a² = 4 and b² = 1
The center is (h,k) = (2,-1)
Plot it:
Since a² = 4, a = 2
Since b² = 1, b = 1
a = 2 is the semi-major axis's length, so draw the
major axis 2a or 4 units long with the center as the midpoint.
We also draw the minor axis 2a or 4 units long with the center
as the midpoint:
Now we can sketch in the ellipse:
The vertices are the endpoints of the major axes,
(0,-1) and (4,-1).
The covertices are the endpoints of the minor
axis, (2,0) and (2,-2).
Edwin
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