Write 36y^2 - 16x^2 = 144 in standard form, and find....

NEHA

Junior Member
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Oct 27, 2006
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I pretty much do all my work my self but, from now on, I'll try to remember to show you all what I am doing.

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Considering the hyperbola, 36y^2 - 16x^2 = 144, write this eqaution in standard form and find the coordinates of the vertices and the equations of the asymototes. Then graph.

. . .36y^2 - 16x^2 = 144

. . .36y^2) / 144 - (16x^2) /144 = 1

I simplified to get:

. . .(y^2) / 4 - (x^2) / 9 = 1

. . .x = 0:
. . .(y^2) /4 - (0)^2 / 9 = 1
. . .y = +/- 2i

. . .y = 0:
. . .(0)^2 / 4 - (x^2) / 9 = 1
. . .x = +/-3

The asymptotes are:

. . .y = (2/3)x and y = -(2/3)x

The coordinates are:

. . .(3, 0) (-3, 0) (0, 2) (0, -2)

Is the above correct? Thank you!
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Edited by stapel -- Reason for edit: spelling, formatting, subject line, etc
 
NEHA said:
I pretty much do all my work my self
That's fine, but we have no way to know this unless you SHOW us.

NEHA said:
from now on I'll show you all what i am doing....
That is what is wanted. Good call. Thank you!

NEHA said:
considering the hyperbola, 36y^2 - 16x^2 = 144...

x = 0
y^2/4 - (0)^2/9 = 1
y = +-2i

y=0
(0)^2/4 - x^2/9 = 1
x = +-3
You seem to be switched around. 'y' is the one with Real solutions, not 'x'.
 
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