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

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]}