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
3 answers
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).
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
2 answers