Question

How do I create an equation of a hyperbola satisfying these conditions: Asymptotes y=3/2x and y=-3/2x; one vertex of (2,0)?

Stuck on this HW question, help much appreciated! thanks.

Answer 1

Find the point of intersection for the asymptotes.

`y = 3/2x`, `y =-3/2x`

Subtract the second equation from the first:

`0 = 6/2x`

`x = 0`

`y = 0`

Because the asymptotes intersect at the point, `(0,0)`, we know that this is the center point of the hyperbola.

Because one of the vertices is 2 horizontal units from the center (at the point `(2,0)`) we know that the hyperbola has a horizontal transverse axis and the general form of its equation is:

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

Substitute center, `(0,0)`, for `(h,k)`:

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

Simplify:

`x^2/a^2-y^2/b^2= 1" [1.1]"`

We can find the value of `a` by substituting the point `(2,0)` into equation [1.1]:

`2^2/a^2-0^2/b^2= 1" [1.1]"`

`2^2/a^2= 1`

`a = 2`

Substitute `a = 2` into equation [1.1]:

`x^2/2^2-y^2/b^2= 1" [1.2]"`

For the horizontal transverse axis type hyperbola, we know that the asymptotes have the general forms:

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

Comparing the first general form with the equation, `y = 3/2x`, we conclude that `b` must equal 3. Substitute `b =3` into equation [1.2]:

`x^2/2^2-y^2/3^2= 1`