Question

How do you graph `12y^2-25x^2=300` and identify the foci and asympototes?

Answer 1

The foci are `(0,3sqrt3)` and `(0,-3sqrt3)`
The asymptotes are `y=5/(2sqrt3)(x)` and `y=-5/(2sqrt3)(x)`

Explanation:

First, we see turn the equation so that it equals one.
`12y^2-25x^2=300 => (12y^2)/300-(25x^2)/300=1`
You can simplify this to:
`(y^2)/25-(x^2)/12=1`

Remember that a hyperbolic equation can be in two forms:
`x^2/a^2-y^2/b^2=1` or `y^2/a^2-x^2/b^2` and `c=sqrt (a^2+b^2)`

Now, a graph of a hyperbola opens to the left and the right when `x^2` is over `a^2`.
Also, a graph of a hyperbola opens to up and down when `y^2` is over `a^2`.

Because of this, the foci of a hyperbola is `(h+-c,k)` when `x^2` is over `a^2`.

Similarly, the foci of a hyperbola is `(h,k+-c)` when `y^2` is over `a^2`.

The equation for the asymptote of a hyperbola is `k+-b/a(x-h)` when `x^2` is over `a^2`.

Similarly, the equation for the asymptote of a hyperbola is `k+-a/b(x-h)` when `y^2` is over `a^2`

Since `y^2` is over `a^2`, we use our appropriate rules to get:
`c=sqrt (12+25) => c=3sqrt3`

Therefore, the foci are `(0,3sqrt3)` and `(0,-3sqrt3)`
The asymptotes are `y=5/(2sqrt3)(x)` and `y=-5/(2sqrt3)(x)`

I have a graph from desmos.com here to visualize it.