Question

How do you find the foci and sketch the hyperbola `x^2/9-y^2/4=1`?

Answer 1

We know that the standard Cartesian form for the equation of a hyperbola with a transverse horizontal axis,

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

has foci at `(h-sqrt(a^2+b^2),k)` and `(h+sqrt(a^2+b^2),k)`

If we write the given equation in the same form as equation [1], then it is a simple matter to find the foci:

`(x - 0)^2/3^2- (y-0)^2/2^2 = 1" [2]"`

Now that we have the given equation in the same form, the computation for the foci is trivial:

`(h-sqrt(a^2+b^2),k)` and `(h+sqrt(a^2+b^2),k)`

`(0-sqrt(9+4),0)` and `(0+sqrt(9+4),0)`

`(-sqrt(13),0)` and `(sqrt(13),0) larr` these are the foci.

To graph the equation you will need the vertices:

`(h-a,k)` and `(h+a,k)`

`(0-3,0)` and `(0+3,0)`

`(-3,0)` and `(3,0) larr` these are the vertices.

And you will need the equations of the asymptotes:

`y = -b/a(x-h) + k` and `y = b/a(x-h) + k`

`y = -2/3(x-0) + 0` and `y = 2/3(x-0) + 0`

`y = -2/3x` and `y = 2/3x larr` there are the equations of the asymptotes.

Here is a graph of, the equation, the foci, the vertices, and the asymptotes.