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

Feb 8, 2017

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

#### Explanation:

Because the foci are horizontally oriented, $\left(- 10 , 0\right) \mathmr{and} \left(10 , 0\right)$, we know that the hyperbola is the horizontal transverse axis type and its equation is:

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

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

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

The reference tells us that the equations of the asymptotes are, $y = - \frac{b}{a} \left(x - h\right) + k \mathmr{and} y = \frac{b}{a} \left(x - h\right) + k$

Matching the above with the given equations, $y = - \frac{3}{4} x \mathmr{and} y = \frac{3}{4} x$, we observe that:

$\frac{b}{a} = \frac{3}{4}$

Which can be written as:

$b = \frac{3}{4} a \text{ [3]}$

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

$10 = \sqrt{{a}^{2} + {b}^{2}} \text{ [4]}$

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

$10 = \sqrt{{a}^{2} + {\left(\frac{3}{4} a\right)}^{2}}$

$100 = {a}^{2} + \frac{9}{16} {a}^{2}$

$100 = \frac{25}{16} {a}^{2}$

${a}^{2} = 64$

$a = 8$

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

$b = \frac{3}{4} \left(8\right)$

$b = 6$

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

${\left(x - 0\right)}^{2} / {8}^{2} - {\left(y - 0\right)}^{2} / {6}^{2} = 1 \text{ [5]}$