# Math Help - Ellipse

1. ## Ellipse

Find the coordinates of the vertices and foci and sketch the graph:

4x^2 + 9y^2 - 32x - 36y + 64 = 0

I tried completing the square and got:

4x^2 - 32x + 256 + 9y^2 - 36y + 324 = 516

but I cant factorise the x and y, since there seem to be no roots

2. Alright. The standard form of an ellipse is:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$

$4x^2 + 9y^2 - 32x - 36y + 64 = 0$

$4x^2 + 9y^2 - 32x - 36y = -64$

Now, group the x's and y's together:

$4x^2 - 32x + 9y^2 - 36y = -64$

Factor out a 4 for the x's, and a 9 for the y's:

$4(x^2-8x) + 9(y^2-4y) = -64$

Now complete the square, for the x's and y's.

$4(x-4)^2 + 9(y-2)^2 = 36$

Since you are completing the square within the parenthesis, you have to multiply by the coefficient of the parenthesis, and that is how I get 36. ( -64 + 64 + 36 = 36)

Next: Divide by 36

$\frac{(x-4)^2}{9} + \frac{(y-2)^2}{4} = 1$

So, now that we have the ellipse in the standard form, we can find the vertices and the foci. The foci of an ellipse =

$c^2 = a^2 - b^2$ Where C is the length of the foci.

Since we know that the foci are on the major axis of the ellipse, we can determine the placing of the foci:

$9 - 4 = 5$

$\sqrt{5}$ is the length of the foci, so now we know that the two foci are at:

$(4\pm\sqrt{5},2)$

The using a and b, we have that the vertices are at:

$(1,2) (7,2)$ along the major axis

$(4,4) (4,0)$ along the minor axis.

If you had some trouble understanding the terminology, or how I came to find the vertices and the foci, I believe this site will be of help to you:

Focus of Ellipse. The formula for the focus and ...