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September 21, 2014

September 21, 2014

Posted by **Maya** on Sunday, April 29, 2012 at 1:20pm.

(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)??

- Pre-calc -
**Reiny**, Sunday, April 29, 2012 at 1:34pmPicking 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**+6y**+9)=36+4**+36**

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

- Pre-calc -
**####Nadine####**, Sunday, April 29, 2012 at 1:36pmWatch 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)

- correction -Pre-calc -
**Reiny**, Sunday, April 29, 2012 at 1:40pmfrom

(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

- Pre-calc -
**Maya**, Sunday, April 29, 2012 at 1:41pmThank you guys so much :)

- Argggh - correction #2Pre-calc -
**Reiny**, Sunday, April 29, 2012 at 1:44pmI 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

- Pre-calc -
**Maya**, Sunday, April 29, 2012 at 1:45pmHaha ok i totally get it thanks!

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