How do you find the vertices, asymptote, foci and graph #x^2/36-y^2=1#?

1 Answer
Douglas K.
May 17, 2017

The General form for a hyperbola of this type is:#(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"#
Vertices: #(h-a,k)# and #(h+a,k)#
Foci: #(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#
Asymptotes: #y=+-b/a(x-h)+k#

Explanation:

Here is a reference for Hyperbola equations provided in the answer section.

Let's convert the given equation, #x^2/36-y^2=1#, to the form of equation [1]:

#(x-0)^2/6^2-(y-0)^2/1^2=1" [2]"#

By inspection of equation [2] we observe that #h = k = 0, a = 6, and b = 1#

Vertices: #(-6,0)# and #(6,0)#
Foci: #(-sqrt(37),0)# and #(sqrt(37),0)#

Asymptotes: #y=1/6x and y = -1/6x#

Here is a graph:

graph{x^2/36-y^2=1 [-10, 10, -5, 5]}

Related questions
See all questions in General Form of the Equation
Impact of this question
2941 views around the world
You can reuse this answer
Creative Commons License