How do you identify the vertices, foci, and direction of #x^2/121-y^2/36=1#?

1 Answer
Apr 10, 2017

Given the equation of a hyperbola with a transverse horizontal axis:
#(x-h)^2/a^2-(y-k)^2/b^2=1#
The vertices are: #(h-a,k)# and #(h+a,k)#
The foci are: #(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#

Explanation:

Note: A horizontal transverse type has a positive x and a negative y term, in a form that is equal to 1, and a vertical transverse type has a negative x term and a positive y term, in a form that is equal to 1.

Given: #x^2/121-y^2/36=1#

Write in the standard form:

#(x-0)^2/11^2-(y-0)^2/6^2=1#

Now, we can write the vertices by observation:

#(-11,0)# and #(11,0)#

Compute #sqrt(a^2+b^2) = sqrt(11^2+6^2)=sqrt(121+36)=sqrt(157)#

Now, we can write the foci with the same ease:

#(-sqrt(157),0)# and #(sqrt(157),0)#