Question

How do you find the standard form of the equation of the hyperbola with the given characteristics? Vertices: (3, 0), (9, 0); foci: (0, 0), (12, 0)

Answer 1

Please observe that the vertices and foci are horizontally oriented, therefore, the standard form is the horizontal transverse axis type:

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

The general form for the vertices of a hyperbola of this type is:

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

The given vertices, `(3, 0) and (9, 0)`, allow us to write 3 equations:

`h-a = 3" [2]"`
`h+a = 9" [3]"`
`k = 0" [4]"`

We can use equations [2] and [3] to solve for `h` and `a`:

`2h = 12`

`h=6" [5]"`

`6+a = 9`

`a = 3" [6]"`

Substitute equations, [4], [5], and [6] into equation [1]:

`(x-6)^2/3^2-(y-0)^2/b^2 = 1" [1.1]"`

The general form for the foci of a hyperbola of this type is:

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

I shall use the focus `(12,0)` to find the value of `b`:

`6+sqrt(3^2+b^2) = 12`

`sqrt(9+b^2) = 6`

`9+b^2= 36`

`b^2= 27`

`b = 3sqrt3" [7]"`

Substitute equation [7] into equation [1.1]:

`(x-6)^2/3^2-(y-0)^2/(3sqrt3)^2 = 1" [1.2]"`

I hope that this helped.