Posted by **Mary** on Monday, January 19, 2009 at 7:22am.

For the ellipse with equation 5x^2+64y^2+30x+128y-211=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)

- Pre-calculus -
**drwls**, Monday, January 19, 2009 at 8:43am
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.

- Pre-calculus -
**tyshiona**, Friday, April 15, 2016 at 11:15am
4x^2-9x+32x-144y-548=0

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