SOLUTION: Find the center of the ellipse 16x^2 + 25y^2 + 32x - 150y = 159

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center of the ellipse 16x^2 + 25y^2 + 32x - 150y = 159      Log On


   



Question 190071: Find the center of the ellipse 16x^2 + 25y^2 + 32x - 150y = 159
Found 2 solutions by stanbon, scott8148:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center of the ellipse 16x^2 + 25y^2 + 32x - 150y = 159
---------------
Complete the square on the x-terms and the y-terms separately:
---
16(x^2 + 2x + ?) + 25(y^2-6y+?) = 159 + ? + ?
16(x^2 + 2x + 1) + 25(y^2 -6y + 9) = 159 +16*1 + 25*9
---
16(x+1)^2 + 25(y-3)^2 = 400
----------------------
Center: (-1,3)
================
Cheers,
Stan H.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
rearranging and factoring ___ 16(x^2+2x) + 25(y^2-6y) = 159

completing the squares ___ 16(x^2+2x+1) + 25(y^2-6y+9) = 159 + 16 + 225 = 400

dividing by 400 ___ {[(x+1)^2] / 25} + {[(y-3)^2] / 16} = 1

this is the "center" form of the equation ___ the center is (-1,3)