Question

How do you graph `y^2-9x^2=9` and identify the foci and asympototes?

Answer 1

The foci are F`=(0,sqrt10)` and F'`(0,-sqrt10)`
The equations of the asymptotes are `y=3x` and `y=-3x`

Explanation:

Let's rewrite the equation as

`y^2/9-x^2/1=1`

This is the equation of an up-down hyperbola.

The general equation is

`(y-k)^2/a^2-(x-h)^2/b^2=1`

The center is `(h,k)=(0,0)`

The vertices are `(h,k+-a)`, A`=(0,3)` and A'`=(0,-3)`

The slope of the asymptotes are `+-a/b` `=+-3`

The equations of the asymptotes are `y=3x` and `y=-3x`

To determine the foci, we need c, `c^2=a^2+b^2`

So, `c=+-sqrt(a^2+b^2)=+- sqrt(9+1)=+-sqrt10`

And the foci are F`=(0,sqrt10)` and F'`(0,-sqrt10)`

graph{(y^2/9-x^2-1)(y-3x)(y+3x)=0 [-12.66, 12.65, -6.32, 6.34]}