How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #(x+6)^2/36-(y+3)^2/9=1#?

1 Answer
Binayaka C.
May 19, 2018

Vertices are at #(-12,-3) and (0,-3) # Focii are at
#((-6-3 sqrt 5),-3) and ((-6+3 sqrt 5),-3) # and
asymptotes are # y= 1/2 x and y= -1/2 x -6#

Explanation:

#(x+6)^2/6^2- (y+3)^2/3^2=1; h=-6,k=-3,a=6 , b= 3#

This is standard form of the equation of a hyperbola with center

#(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :. # Center #(-6,-3)#

The vertices are #a# units from the center, and the foci are

#c# units from the center. Moreover #c^2=a^2+b^2# or

#c^2=6^2+3^2 :. c^2=45 or c = +- 3sqrt5#

Vertices are at #(-12,-3) and (0,-3) #

Focii are at #((-6-3 sqrt 5),-3) and ((-6+3 sqrt 5),-3) #

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions #2 a=12 and 2 b=6 # with its

center at #(-6,-3) :.# slope #=+-b/a=+- 3/6= +- 1/2#.

Equation of asymptotes are #y+3= +-1/2(x+6) # or

#y=-3+-1/2(x+6) or y= 1/2 x and y= -1/2 x -6#

Therefore, asymptotes are # y= 1/2 x and y= -1/2 x -6#

graph{(x+6)^2/36-(y+3)^2/9=1 [-80, 80, -40, 40]}

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