Question

4x^2-9y^2-40x+36y+100=0


I still end up with the wrong answer. I end up with
4(x-5)^2/-120 - 9(y+2)^2=-120

9 can't go into -120

My procedure:
(4x^2-40x)-(9y^2+36y)=-100
4(x^2-10x +4)-9(y^2 +4 +4)=-100+ 4(4)-9(4)
4(x-5)^2-9(y+2)^2=-120
(x-5)^2/-30 - ? 9 can't go into -120 evenly

Answer 1

4x^2-9y^2-40x+36y+100=0

4x^2-40 x - 9y^2+36y = -100


4x^2-40 x - 9y^2+36y = -100

x^2 -10x - (9/4)y^2 +9 y = -25

x^2 -10x +25 - (9/4)y^2 +9 y = 0

(x-5)^2 - (9/4)y^2 +9 y = 0

4(x-5)^2 - 9y^2 +36 y = 0

(4/9)(x-5)^2 - y^2 +4 y = 0

(4/9)(x-5)^2 - y^2 +4 y +4 = 4

(4/9)(x-5)^2 - (y+2)^2 = 4

4(x-5)^2 - 9(y+2)^2 = 36

(1/36)4(x-5)^2-(1/36)9(y+2)^2 = (1/36)36

(1/9)(x-5)^2 - (1/4)(y+2)^2 = 1

Answer 2

(4/9)(x-5)^2 - y^2 +4 y = 0

-(4/9)(x-5)^2 + y^2 -4 y = 0

-(4/9)(x-5)^2 + y^2 -4 y + 4 = 4

-(4/9)(x-5)^2 + (y-2)^2 = 4

-4(x-5)^2 + 9(y-2)^2 = 36

-(1/9)(x-5)^2 + (1/4)(y-2)^2 = 1