Precalculus question

Question

From the equation of the ellipse 16x² + 9y² + 160x - 126y + 697 = 0 , find the following:

1. Coordinates of the center
2. Coordinates of the two focal points
3. Length of the major axis
4. Length of the minor axis

Answer 1

16(x^2 + 10x + 25) + 9(y^2 -14y + 49) = 400 + 441 -697

complete the squares, similar to solving a quadratic equation

4^2(x+5)^2 + 3^2(y-7)^2 = 144 = (12)^2

(x+5)^2/(12/4)^2 + (y-7)^2/(12/3)^2 = 1

simplify

(x+5)^2/3^2 + (y-7)^2/4^2 = 1

center is (-5,7)

axes are 4 and 3

foci or the 2 focuses are points where the distance from the ellipse to them added together is always the same. the foci are on the major axis, so it's (-5,7+f) and (-5,7-f) where 7 + or - f are the y-coordinates of the foci.

take the point (-5,11) and calculate it's distance to each focus and add them. 11-7-f + 11-7+f = 22-14 = 8

take the point (-2,7) and calculate its distance to each focus and add them, then set that = 8

then solve for f = sqr7

the foci are (-5,7+sqr7) and (-5,7-sqr7)