How do you find all the critical points to graph #x^2 - y^2 + 9 = 0# including vertices, foci and asymptotes?

1 Answer
Jul 9, 2018

Answer:

Center: #(0,0)#, vertices are at #(0,3) and (0,-3)#, focii are at#(0,3 sqrt 2) and (0,-3 sqrt 2)#, equation of asymptotes
of vertical hyperbola are
#y=+-x#

Explanation:

# x^2-y^2+9=0 or y^2- x^2=9# or

#y^2/3^2- x^2/3^2=1#.The standard equation of vertical

hyperbola is #(y-k)^2/a^2-(x-h)^2/b^2=1; h,k# being center.

Here # h=0 ,k=0 , a=3 , b=3#. So center is at # (0,0)#

Vertices are #a=3# units from center . Therefore two vertices

are at #(0,3) and (0,-3)# , Focii are #c# units from center.

#c^2=a^2+b^2=9+9=18 :. c= 3 sqrt2#. Therefore focii are

#(0,3 sqrt 2) and (0,-3 sqrt 2)#. 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=6 and 2 b=6 # with its center at #(0,0) :.#

slope #+-a/b=+- 1#. Equation of asymptotes are

#y-0= +-1(x-0) or y=+-x#

graph{x^2-y^2+9=0 [-10, 10, -5, 5]}[Ans]