How do you find the center and vertices and sketch the hyperbola #4y^2-9x^2=36#?

1 Answer
Jun 27, 2017

Please see below.

Explanation:

#4y^2-9x^2=36# is the equation of a vertical hyperbola, wherein center, foci and vertices line up above and below each other, parallel to the #y#-axis.

The general form of such an equation is #(y-k)^2/a^2+(x-h)^2/b^2=1#

where #(h,k)# is the center and vertices are #(h,k+a)# and #(h,k-a)# and equation of asymptotes are #y=+-a/b(x-h)+k#.

as #4y^2-9x^2=36# can be written as #(y-0)^2/3^2-(x-0)^2/2^2=1# and as such

center is #(0,0)# and vertices are #(0,3)# and #(0,-3)#, while asymtotes are #3x-2y=0# and #3x+2y=0#.

The graph appears as shown below.

graph{(4y^2-9x^2-36)(3x-2y)(3x+2y)=0 [-20, 20, -10, 10]}