Finding the center of an Ellipse?

[Cryptologica]

Homework Statement

The ellipse 18x^2+2x+y^2=1 has its center at the point (b,c) where b=____ and c=____?

Homework Equations

x^2/a^2 + y^2/b^2 = 1

The Attempt at a Solution

18x^2+2x+y^2=1
18(x^2+(1/9)x)+y^2=1
18(x^2+(1/9)x+(1/324))+y^2= 1+18(1/324)
18(x+(1/18))^2+y^2=19/18

I need it to be in the form x^2/a^2 + y^2/b^2 = 1, then I can easily determine the coordinates of the center. Did I do something wrong? How do I get to this form? Thanks.
(This is for my Calc. II class and I asked in two other forums and got nothing...but I figured you guys would know about this due to applications in astrophysics.)

 

[Simon Bridge]

You didn't do anything wrong, you are almost there.

The equation you are looking for is more like:

\frac{(x-x_0)^2}{a}+\frac{(y-y_0)^2}{b}=1

... where a and b are the semi-axis, and the center is at (x_0,y_0) (and the ellipse has not been rotated.) Notice you can just read the center off from your equation.

 

[Cryptologica]

You didn't do anything wrong, you are almost there.

The equation you are looking for is more like:

\frac{(x-x_0)^2}{a}+\frac{(y-y_0)^2}{b}=1

... where a and b are the semi-axis, and the center is at (x_0,y_0) (and the ellipse has not been rotated.) Notice you can just read the center off from your equation.

Yes, I know I am really close. However, I cannot get the equation to equal one, without having constants in the numerator other than 1. When I multiply by (18/19) to make it equal to one the numerator never cancels?

 

[LCKurtz]

Yes, I know I am really close. However, I cannot get the equation to equal one, without having constants in the numerator other than 1. When I multiply by (18/19) to make it equal to one the numerator never cancels?

You can always get a constant out of the numerator. Remember that\frac a b = \frac 1 {\frac b a}

 

[Cryptologica]

Oh, ok! Thanks, I think I got it now.