Math
posted by Anonymous on .
9x^2+16y^218x+64y71=0
find the coordinates of the center, the foci, and the vertices of this ellipse.
9x^2+16y^218x+64y71=0
9x^218x+16y^2+64y=71
9(x^22x)+16(y^2+4y)=71
9(x^22x+1)+16(y^2+4y+4)=71

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.

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). 
If the exterior angle of a regular polygon is 45o, then find the number of sides of the polygon