Find the center of an ellipse?

Question

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

Answer 1

First, express your ellipse in standard form:

( x - h )^2 / a^2  + ( y - k )^2 / b^2  = 1

where

a = length of semimajor axis

b = length of semiminor axis

( h , k ) are the coordinates of the center

b^2  ( x - h )^2  + a^2 ( y - k )^2   = (ab)^2

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

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

3^2 ( x^2 - 2x ) + 4^2 ( y^2 + 4y ) = 71

3^2 ( x - 1 )^2 - 9 + 4^2 ( y + 2 )^2 - 64 = 71

3^2 ( x - 1 )^2 + 4^2 ( y + 2 )^2 = 144

So the parameters are a = 3, b = 4, h = 1, and k = -4.

The center is at x = h, y = k.