Finding the center of an Ellipse?

  • #1

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

  • #2
Simon Bridge
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You didn't do anything wrong, you are almost there.

The equation you are looking for is more like:

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

... where a and b are the semi-axis, and the center is at [itex](x_0,y_0)[/itex] (and the ellipse has not been rotated.) Notice you can just read the center off from your equation.
 
  • #3
You didn't do anything wrong, you are almost there.

The equation you are looking for is more like:

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

... where a and b are the semi-axis, and the center is at [itex](x_0,y_0)[/itex] (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?
 
  • #4
LCKurtz
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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[tex]\frac a b = \frac 1 {\frac b a}[/tex]
 
  • #5
Oh, ok! Thanks, I think I got it now.
 

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