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
-----------------------------------------
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!
____________________________________________
Edited by stapel -- Reason for edit: spelling, formatting, subject line, etc