Posted by **Anonymous** on Tuesday, August 26, 2008 at 6:21pm.

9x^2+16y^2-18x+64y-71=0

find the coordinates of the center, the foci, and the vertices of this ellipse.

9x^2+16y^2-18x+64y-71=0

9x^2-18x+16y^2+64y=71

9(x^2-2x)+16(y^2+4y)=71

9(x^2-2x+1)+16(y^2+4y+4)=71

- Math -
**bobpursley**, Tuesday, August 26, 2008 at 7:00pm
The last line is just wrong. You added 9 +64 to the left, and did nothing to the right. If you add things to one side, you have to add to the other.

- Math -
**Anonymous**, Tuesday, August 26, 2008 at 7:14pm
9x^2 - 18x + 16y^2 + 64y = 71

9(x^2 - 2x) + 16(y^2 + 4y) = 71

9(x^2- 2x + 1) + 16(y^2 + 4y + 4) = 71 + 9 + 64

9(x - 1)^2 + 16(y + 2)^2 = 144

(x - 1)^2/16 + (y + 2)^2/9 = 1

centered at (1, -1), major axis length 8, minor axis length 6.

Vertices are at (5, -1) and (-3, -1).

- Math -
**shankeria**, Sunday, April 18, 2010 at 7:53pm
If the exterior angle of a regular polygon is 45o, then find the number of sides of the polygon

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