Question

How do you find the equation of the hyperbola with center at the origin and foci `(+-10,0)` and asymptotes `y=+-3/4x`?

Answer 1

I am going to use this reference to answer this question.

Explanation:

Because the foci are horizontally oriented, `(-10, 0) and (10,0)`, we know that the hyperbola is the horizontal transverse axis type and its equation is:

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

We are told that the center is at the origin, therefore, both h and k equal zero; this makes equation [1] become equation [2]:

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

The reference tells us that the equations of the asymptotes are, `y = -b/a(x - h) + k and y = b/a(x - h) + k`

Matching the above with the given equations, `y = -3/4x and y = 3/4x`, we observe that:

`b/a = 3/4`

Which can be written as:

`b = 3/4a" [3]"`

Because the foci are a distance of 10 from the origin, we can write the following equation:

`10 = sqrt(a^2 + b^2)" [4]"`

Substitute equation [3] into equation [4] and then solve for "a":

`10 = sqrt(a^2 + (3/4a)^2)`

`100 = a^2 + 9/16a^2`

`100 = 25/16a^2`

`a^2 = 64`

`a = 8`

Use equation [3] to compute the value of b:

`b = 3/4(8)`

`b = 6`

This makes equation [2] become equation [5]:

`(x - 0)^2/8^2 - (y - 0)^2/6^2 = 1" [5]"`

Equation [5] is the answer.