How do you graph #36x^2-4y^2=144# and identify the foci and asympototes?

1 Answer
Nov 10, 2016

Answer:

The foci are F#=(sqrt40,0)# and F'#(-sqrt40,0)#
The equations of the asymptotes are #y=3x# and #y=-3x#

Explanation:

The equation represents a hyperbola.
#x^2/4-y^2/36=1#
The general equation is
#(x-h)^2/a^2-(y-k)^2/b^2=1#
The center is #(0,0)#

The vertices are #(h+a,k)# and #(h-a,k)#
The vertices are #(2,0)# and #(-2,0)#

The slope of the asymptotes are #+-b/a=+-6/2=+-3#
The equations of the asymptotes are #y=k+-b/a(x-h)#
#y=3x# and #y=-3x#
To determine the foci, we need #c=+-sqrt(a^2+b^2)#

#c=+-sqrt(4+36)=+-sqrt40#
the foci are #(h+-c,k)#
the foci are F#=(sqrt40,0)# and F'#(-sqrt40,0)#
graph{(x^2/4-y^2/36-1)(y+3x)(y-3x)=0 [-18.02, 18.03, -9, 9.02]}