Question

How do you find the standard form of the equation of the hyperbola given the properties foci `(+-5,0)`, length of the conjugate axis 6?

Answer 1

Please see the explanation.

Explanation:

Because the foci are given to be `(-5,0)` and `(5,0)`, we know that the hyperbola is the horizontal transverse axis type and has the general equation:

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

We know that `2b` is the length of the conjugate axis; this allows us to write the equation:

`2b = 6`

`b=3`

Substitute the value for "b" into equation [1]:

`(x-h)^2/a^2-(y-k)^2/3^2=1" [2]"`

We know that the general forms for the foci are:

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

This allows us to write the following equations:

`k=0" [3]"`
`-5 = h-sqrt(a^2+3^2)" [4]"`
`5 = h+sqrt(a^2+3^2)" [5]"`

Substitute the value for k into equation [2]:

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

Find the value of h by adding equations [4] and [5]:

`2h = 0`

`h = 0`

Substitute into equation [6]:

`(x-0)^2/a^2-(y-0)^2/3^2=1" [7]"`

Use equation [5] to find the value of "a":

`5 = 0+sqrt(a^2+3^2)`

#25 = a^2+9

`a^2 = 16`

`a = 4`

Substitute the value of "a" into equation [7]

`(x-0)^2/4^2-(y-0)^2/3^2=1" [8]"`

Equation [8] is the equation of the hyperbola.