Question

How do you find the standard form of the equation of the hyperbola given the properties vertices (-10,5), asymptotes `y=+-1/2(x-6)+5`?

Answer 1

There are two standard forms:

1. `(x - h)^2/a^2 -(y-k)^2/b^2 = 1`

2. `(y - k)^2/a^2 -(x-h)^2/b^2 = 1`

The point-slope form for the equations of the asymptotes is:

`y = +-m(x-h)+k`

Therefore, the equations, `y=+-1/2(x-6)+5` tell us that `h = 6` and `k =5` -- i. e. The center is the point `(6,5)`.

This information coupled with one of the vertices given to be, `(-10,5)`, tells us that the hyperbola is the horizontal transverse type with the equation in item 1:

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

Substitute 6 for h and 5 for k:

`(x - 6)^2/a^2 -(y-5)^2/b^2 = 1" [1.1]"`

The point `(-10,5)` allows us to discover the value of "a":

`(-10 - 6)^2/a^2 -(5-5)^2/b^2 = 1`

`(-16)^2/a^2 -(0)^2/b^2 = 1`

`256/a^2 = 1`

`256 = a^2`

`a = 16`

Substitute the value of "a" into equation [1.1]:

`(x - 6)^2/16^2 -(y-5)^2/b^2 = 1" [1.2]"`

The general form for the asymptotes of a hyperbola with a horizontal transverse axis is:

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

Again, using the equations, `y=+-1/2(x-6)+5`, we can write the following equation:

`b/a = 1/2`

Substitute 16 for "a" and solve for "b":

`b = 16/2`

`b = 8`

Substitute the value for "b" into equation [1.2]:

`(x - 6)^2/16^2 -(y-5)^2/8^2 = 1" [1.3]"`