Question

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`?

Answer 1

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]}