Need help getting equation in ellipitcal format?

Question

12x^2 + 7y^2 - 120x - 28y + 307 = 0



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

Can someone show me how they get from the 1st equation to the 2nd so i can graph this ellipse?

Answer 1

You want to gather terms and complete squares.

12x² + 7y² - 120x - 28y + 307 = 0
12x² - 120x + 7y² - 28y + 307 = 0
12(x² - 10x + 25) + 7(y² - 4y + 4) + 307 - 12*25 - 7*4 = 0
12(x - 5)² + 7(y - 2)² + (307 - 300 - 28) = 0
12(x - 5)² + 7(y - 2)² = 21
12(x - 5)²/21 + 7(y - 2)²/21 = 1
(x - 5)² / (21/12) + (y - 2)² / (21/7) = 1
(x - 5)² / (7/4) + (y - 2)² / 3 = 1

Answer 2

group x terms and y terms:
12x² - 120x + 7y² - 28y = -307

factor out each quadratic coefficient:
12(x² - 10x ....) + 7(y² - 4y ....) = -307

complete each square:
12(x² - 10x + 25) + 7(y² - 4y + 4) = -307 + 12•25 + 7•4
12(x - 5)² + 7(y - 2)² = 21

divide by 21:
(4/7)(x - 5)² + (1/3)(y - 2)² = 1

final rewrite:
(x - 5)² / (7/4) + (y - 2)² / 3 = 1