Question

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

Answer 1

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]}