Question regarding hyperbolas?

Question

What is the vertices, asymptotes and foci of this hyperbola?
36x^2-4y^2=4??

Please show me the step to reach the answer!

Points will be awarded ASAP!!!!

Answer 1

first you have to put the hyperbola in standard form. Divide everything by 4.

9x^2 - y^2 = 1
Which is the same as
(x^2)/(1/9) - (y^2)/1 = 1
Here, a^2 = 1/9, so a = 1/3,
and b^2 = 1, so b = 1

In this case, the center is (0,0)
Since the "x" term is first, then it opens left and right. The vertices will be left and right from the center by "a" units, so the vertices are (1/3, 0) and (-1/3, 0).

The foci are left and right of the center by "c" units. In a hyperbola, a^2 + b^2 = c^2, so c^2 = (1/3)^2 + (1^2) = 1/9 + 1 = 10/9. c = root(10)/3. The foci will be (root10/3, 0) and (-root10/3, 0).

The asymptotes have slopes that are +b/a and -b/a, and pass through the center, in this case, (0,0). So these slopes are +1/(1/3) and -1/(1/3) or 3 and -3, and the equations are y = 3x and y = -3x.

Answer 2

36x^2-4y^2=4
Divide by 4: x^2/9-y^2=1
So we have a=3 and b=1
Since this is a horizontal hyperbola, and the center is at the origin, the vertices are (-3,0) and (3,0).

To get the foci:
c^2=a^2+b^2 in a hyperbola so,
c^2=9+1
c^2=10
c=square root of 10
so the foci are (-squareroot of 10,0) and (squareroot of 10,0)

Asymptotes:
y=(b/a)x and y=-(b/a)x (in a horizontal hyperbola)
so the asymptotes are:
y=1/3x and y=-1/3x

Answer 3

Check these two website out:

http://documents.wolfram.com/teachersedition/Teacher/Hyperbolas.html

http://mathworld.wolfram.com/Hyperbola.html

Answer 4

just look to wikipedia