Question

Heres the equation:
(x^2)-(4y^2)-(4x)+(24y)-(36)=0
Your supposed to find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of its graph.
So first you complete the square right?
(x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
[((x+2)^2)/76] -[(y+3)^2)/19]=1
So its a hyperbola, and is the center at (-2, -3)??

Answer 1

Picking it up from your

(x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
should have been
(x^2 -4x+4)-4(y^2 <b>+6y</b>+9)=36+4<b>+36</b>
notice on the left you had
... 4(..... +9) , so you subtracted 36, thus you must subtract 36 from the right side as well
(you had added 36)

final version

(x-2)^2 - 4(y-3)^2 = 36

or

(x-2)^2 /36 - (y-3)^2 /9 = 1

centre would be (2,3)
a = 1 , b = 3 , c = √10

vertices: (1,3) and (3,3)
foci :( 2-√10 , 3) and (2+√10 , 3)
asymptotes:

first: y = 3x + b , with (2,3) on it
3 = 6+b
b = -3 ----> y = 3x - 3
second: y = -3x + b
3 = -6+b
b = 9 -----> y = -3x + 9

Answer 2

Watch your signs!
(x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
should be (x^2 -4x+4)-4(y^2 -6y+9)=36+4-36
(x^2 -4x+4)/4-4(y^2 -6y+9)/4=4/4
((x-2)^2)/4- (y-3)^2 =1
hyperbola, center at (2,3)

Answer 3

from
(x-2)^2 /36 - (y-3)^2 /9 = 1 , the rest should say

centre is (2,3)
a = 6, b = 3, c = √45


vertices: (-4,3) and (8,3)
foci :( 2-√45 , 3) and (2+√45 , 3)
asymptotes:

first: y = (1/2)x + b , with (2,3) on it
3 = 1+b
b = 2 ----> y = (1/2)x + 2
second: y = -(1/2)x + b
3 = 1+b
b = 2 -----> y = -(1/2)x + 2

check my arithmetic

Answer 4

Thank you guys so much :)

Answer 5

I am sorry, don't know where I got that -4 on the RS from
Go with Nadine's final equation
(x^2 -4x+4)-4(y^2 -6y+9)=36+4-36

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

from there, a=2, b = 1, c = √5

adjust from there

Answer 6

Haha ok i totally get it thanks!