How do you graph # (y+1)^2/36 - (x+5)^2/9 =1#?

1 Answer
May 24, 2017

Answer:

#"center " = (-5, -1); " "a = 6; " "b = 3#
#"vertices "(-5, 5), (-5, -7)#

Explanation:

Given: #(y+1)^2/36 - (x+5)^2/9 = 1#

Form is a Vertical Transverse axis hyperbola: #(y-k)^2/a^2 - (x-h)^2/b^2 = 1#

#"center " = (-5, -1)#

#a = sqrt(36) = 6; " " b = sqrt(9) = 3#

Since the larger number in the denominator is over the #y# value, this is the #a#. Create a dashed box that is #a = 6# above and below the center in the #y# direction. Make it be #b = 3# to the right and left of the center.

Create dashed lines that connect the corners (asymptotes) and extend past the corners. The equations of the asymptotes are: #y = +- a/b(x-h)+k# for the vertical transverse hyperbola.

Let the vertices be at #(h, k+-6) = (-5, 5), (-5, -7)#

The hyperbola has two branches. Each branch goes through one vertex and stays inside the asymptotes.

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