Question

How do you find the center, vertices, foci and asymptotes of `x^2/7 - y^2/9=1`?

Answer 1

The graph should look like this:
graph{x^2/7-y^2/9=1 [-10, 10, -5, 5]}

Explanation:

`x^2/7-y^2/9=1`

Since we are subtracting, we know that this is a hyperbola. Also, since the fraction with `x^2` is positive, this hyperbola opens to the right and the left.
Therefore, equation is in the form `x^2/a^2-y^2/b^2`
We know that:
`h=0`
`k=0`
`a=sqrt7`
`b=sqrt9` or `3`.
`c=sqrt (a^2+b^2)` or `4`.
Knowing this, we already know which equations to use:
The center is always `(h,k)`
The vertices are `(h+a,k)` and `(h-a,k)`
The foci are `(h+c,k)` and `(h-c,k)`
Since this hyperbola opens sideways, the asymptotes are found by this formula:
`y=k+-b/a(x-h)`

Therefore, we know that the center is `(0,0)`
The vertices are approximately `(2.645,0)` and `(-2.645,0)`
The foci are `(4,0)` and `(-4,0)`
The lines of asymptotes are `y=(3sqrt7)/7x` and `y=-(3sqrt7)/7x`