How do you find the equation of the hyperbola with center at the origin and vertices (+-1,0) and asymptotes y=+-3x?

Jan 2, 2017

Explanation:

Please observe that the vertices are horizontally oriented, $\left(- 1 , 0\right) \mathmr{and} \left(1 , 0\right)$, therefore, the hyperbola is the horizontal transverse axis type . From the reference, the standard Cartesian form for the equation of a hyperbola with a horizontal transverse axis is:

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

where h and k are the center point $\left(h , k\right)$, "a" is the distance from the center to a vertex, and $\pm \frac{b}{a}$ is the slope of asymptotes.

We are given that the center is the origin so substitute 0 for h and k into equation [1]:

${\left(x - 0\right)}^{2} / {a}^{2} - {\left(y - 0\right)}^{2} / {b}^{2} = 1 \text{ [2]}$

We can see that the distance from the center to either vertex is 1, therefore substitute 1 for a into equation [2]:

${\left(x - 0\right)}^{2} / {1}^{2} - {\left(y - 0\right)}^{2} / {b}^{2} = 1 \text{ [3]}$

We are given that the slope of the vertices is $\pm 3$; this allows us to solve for b:

$\pm \frac{b}{a} = \pm 3$

Substitute 1 for a:

$\pm \frac{b}{1} = \pm 3$

$b = 3$

To substitute 3 for b into equation [3]:

${\left(x - 0\right)}^{2} / {1}^{2} - {\left(y - 0\right)}^{2} / {3}^{2} = 1 \text{ [4]}$ Answer

You can simplify equation [4] if you like, but I recommend that you leave it in this form.