Hyperbola equation: What went wrong?

Question

I need to know where I messed up and how to get the right answers for the center, foci, major & minor axes.
The Original Equation:
36x^2 - 100y^2 - 72x + 400y = 3964
Factored:
18(2x^2 - 4x) - 100(y^2 + 4y) = 3964
After I add (-4/2)^2 and (4/2)^2 to the left side and 36 & -200 to the left:
18(2x^2 - 4x + 2) - 100(y^2 + 4y + 2) = 3964 + 36 -200
After I divided by 3800 on both sides:

(x-2)^2 - (y+2) = 1
----------- -----------
100 9

Last Format:

(x-2)^2 - (y - (-2))^2
---------- -------------- = 1
10^2 3^2

So I got a=10 b=3 h=2 and k= -2
So for the center = (2,-2)
Foci Points = (2-sqrt(109), -2) & (2+sqrt(109), -2)
and the major and minor axises: 20, minor=6

WHERE DID I GO WRONG???

I realize it seems like a lot, but it would mean so much. I finally thought I was getting math and everything "went wrong". I overlooked it myself more then 10 times and didn't catch anything. Please, I beg of you, explain what I did. Thank you very much!!!

Merry CHRISTmas!

***Sorry about the little things that may be hard to understand, when I posted my original format changed a bit. The denominators moved over in both parts. Thanks in advance guys!

Answer 1

Find the center, foci, and major and minor axes of the hyperbola.

36x² - 100y² - 72x + 400y = 3964

(36x² - 72x) - (100y² - 400y) = 3964
36(x² - 2) - 100(y² - 4y) = 3964

This is the first place you went wrong. You wrote
- 100(y^2 + 4y) instead of - 100(y^2 - 4y). A minus times a minus is a plus. For the x terms it is better to factor out 36 so that the coefficient for the x² term is a perfect square.

36(x² - 2x + 1) - 100(y² - 4y + 4) = 3964 + 36*1 - 100*4
You made a mistake here. You had
-100*2 instead of -100*4

36(x - 1)² - 100(y² - 2)² = 3600
You made a mistake here. You had (x - 2) instead of (x - 1).

(x - 1)²/100 - (y² - 2)²/36 = 1

a² = 100
a = 10

b² = 36
b = 6

c² = a² + b² = 100 + 36 = 136
c = √136 = 4√34
_________

The center of the hyperbola is (h, k) = (1, 2).

Since the x term is positive and the y term negative, the hyperbolas open sideways.

The foci are located at:
(h ± c, k) = (1 ± 4√34, 2)

The vertices are located at:
(h ± a, k) = (1 ± 10, 2) = (-9, 2) and (11, 2).

The transverse axis is the distance between the two vertices. It is 2a = 20.
(For hyperbolas we call this the transverse axis rather than the major axis like we do for elllipses, since unlike ellipses, it can be either smaller or larger than the other axis.)

The conjugate axis is the axis at right angles to the transverse axis.
It is 2b = 12.
(For hyperbolas we call this the conjugate axis rather than the minor axis like we do for elllipses, since unlike ellipses, it can be either smaller or larger than the transverse axis.)

Answer 2

36x^2 - 100y^2 - 72x + 400y = 3964
36(x^2 -2x) -100(y^2 -4y) = 3964
36(x-1)^2 -36 -100(y -2)^2 +400 = 3964
36(x-1)^2 -100(y-2)^2 = 3600

*******************
Completing the square, you forgot that you had 2x^2, you thought you had x^2

Answer 3

Merry Christmas to you....you got CARELESS...what is (4/2)^2 = ???...after felling foolish rework the problem...your analysis was excellent!!!