How do you find all the critical points to graph #x^2 /9- y^2 /9 =1 # including vertices, foci and asymptotes?

1 Answer
Nov 4, 2016

Answer:

The vertices are #( +3,0)# and #(-3,0)#
The foci are #F=(3sqrt2,0)# and #F'=(-3sqrt2,0)#
The asymptotes are #y=x# and #y=-x#

Explanation:

The equation is a left right hyperbola.
#(x-h)^2/a^2-(y-k)^2/b^2=1#
The center is #h,k##=>##(0,0)#
The vertices are #(h+-a,k)##=>#( +3,0)# and #(-3,0)#

The slopes of the asymptotes are #+-b/a##=+-3/3=+-1#
The equations of the asymptotes are #y=k+-b/a(x-h)#
#:. y=+-x#
To calculate the foci, we need c, #c^2=a^2+b^2#
#:. c=+-sqrt18#
and the foci are #F=(3sqrt2,0)# and #F'=(-3sqrt2,0)#

graph{x^2/9-y^2/9=1 [-12.66, 12.65, -6.33, 6.33]}