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
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
(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
It seems like you made a small mistake in your procedure. Let me explain how to correctly manipulate the equation.
Starting with the equation:
4x^2 - 9y^2 - 40x + 36y + 100 = 0
1. Group the x-terms and the y-terms separately:
(4x^2 - 40x) - (9y^2 - 36y) + 100 = 0
2. Factor out common factors from each group (x and y terms):
4(x^2 - 10x) - 9(y^2 - 4y) + 100 = 0
3. Complete the square for both the x and y groups:
4(x^2 - 10x + 25 - 25) - 9(y^2 - 4y + 4 - 4) + 100 = 0
4. Simplify within each group:
4((x-5)^2 - 25) - 9((y-2)^2 - 4) + 100 = 0
5. Distribute the constants within each group:
4(x-5)^2 - 100 - 9(y-2)^2 + 36 + 100 = 0
6. Combine like terms:
4(x-5)^2 - 9(y-2)^2 + 36 = 0
Now, let's check the mistake you mentioned. You're correct that 9 cannot go into -120 evenly. However, the equation we obtained in step 6 does not have a constant term on the right-hand side. That means the negative constant term (-120 in your previous equation) should be on the left-hand side. The equation should be:
4(x-5)^2 - 9(y-2)^2 = -36
So, the correct equation is:
4(x-5)^2 - 9(y-2)^2 = -36
I hope this clears up the confusion. Let me know if you have any further questions!