How do you find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph of #x^2/9 - y^2/16 = 1#?

1 Answer
Mar 4, 2017

Answer:

The center is #(0,0)#
The vertices are (-3,0)# and #(3,0)#

The foci are #F'=(-5,0)# and #F=(5,0)#
The asymptotes are #y=4/3x# and #y=-4/3x#

Explanation:

We compare this equation

#x^2/3^2-y^2/4^2=1#

to

#x^2/a^2-y^2/b^2=1#

The center is #C=(0,0)#

The vertices are #V'=(-a,0)=(-3,0)# and #V=(a,0)=(3,0)#

To find the foci, we need the distance from the center to the foci

#c^2=a^2+b^2=9+16=25#

#c=+-5#

The foci are #F'=(-c,0)=(-5,0)# and #F=(c,0)=(5,0)#

The asymptotes are

#x^2/3^2-y^2/4^2=0#

#y=+-4/3x#

graph{((x^2)/9-(y^2)/16-1)(y-4/3x)(y+4/3x)=0 [-16.02, 16.02, -8.01, 8.01]}