# 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?

Oct 30, 2017

There are two standard forms:

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

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

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

$y = \pm m \left(x - h\right) + k$

Therefore, the equations, $y = \pm \frac{1}{2} \left(x - 6\right) + 5$ tell us that $h = 6$ and $k = 5$ -- i. e. The center is the point $\left(6 , 5\right)$.

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

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

Substitute 6 for h and 5 for k:

${\left(x - 6\right)}^{2} / {a}^{2} - {\left(y - 5\right)}^{2} / {b}^{2} = 1 \text{ [1.1]}$

The point $\left(- 10 , 5\right)$ allows us to discover the value of "a":

${\left(- 10 - 6\right)}^{2} / {a}^{2} - {\left(5 - 5\right)}^{2} / {b}^{2} = 1$

${\left(- 16\right)}^{2} / {a}^{2} - {\left(0\right)}^{2} / {b}^{2} = 1$

$\frac{256}{a} ^ 2 = 1$

$256 = {a}^{2}$

$a = 16$

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

${\left(x - 6\right)}^{2} / {16}^{2} - {\left(y - 5\right)}^{2} / {b}^{2} = 1 \text{ [1.2]}$

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

$y = \pm \frac{b}{a} \left(x - h\right) + k$

Again, using the equations, $y = \pm \frac{1}{2} \left(x - 6\right) + 5$, we can write the following equation:

$\frac{b}{a} = \frac{1}{2}$

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

$b = \frac{16}{2}$

$b = 8$

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

${\left(x - 6\right)}^{2} / {16}^{2} - {\left(y - 5\right)}^{2} / {8}^{2} = 1 \text{ [1.3]}$