# 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.)

## Answers and Replies

Homework Helper
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?

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