Precalculus
posted by Mary .
For the ellipse with equation 5x^2+64y^2+30x+128y211=0, find the cooridinates of the center, foci, and vertices. Then, graph the equation.
my answer is:
coordinates of center: (3,1)
foci:(11,1) and (4.7,1)
vertices: (11,1) (5,1)
(3,1+square root of 5)
not sure how to graph, i know the center is (3,1)

5(x^2 + 6x + 9) + 64(y^2 + 2x + 1) 45 64 211 = 0
5(x+3)^2 + 64 (y+1)^2 = 320
(x+3)^2/64 + (y+1)^2/5 = 0
Shouldn't the center be at (3, 1) ?
The semimajor axis length (along the x direction) is a = sqrt64 = 8 and the semiminor axis length is b = sqrt5. The foci are at y = 1, and x = 3 +/ c, where
c^2 = a^2  b^2 = 64  5 = 59
Add and subtract the semimajor axis lengths from the center coordinate to get the vertex locations.
I can't help you with the graphing part. You will need to locate the center on a graph, plot the vertex locations, and compute the coordinates of some intermediate points along the curve.
It is possible that I have made some math errors myself, so check my work. 
4x^29x+32x144y548=0