How do you find the standard form of the equation of the hyperbola given the properties vertex (0,1), vertex (8,1), focus (-3,1)?

Aug 1, 2017

${\left(x - 4\right)}^{2} / 16 - {\left(y - 1\right)}^{2} / 33 = 1$

Explanation:

As the two vertex are $\left(0 , 1\right)$ and $\left(8 , 1\right)$ and focus is $\left(- 3 , 1\right)$

as the distance between $\left(0 , 1\right)$ and $\left(- 3 , 1\right)$ is $3$, the other focus is $\left(11 , 1\right)$ (at a distance of $3$ to the right of $\left(8 , 1\right)$) and central point is $\left(4 , 1\right)$.

Hence, $a = 4$, the distance to either side of center and $c = 7$, the distance from center to focus.

Hence, ${b}^{2} = {c}^{2} - {a}^{2} = 33$

and equation is

${\left(x - 4\right)}^{2} / 16 - {\left(y - 1\right)}^{2} / 33 = 1$

graph{((x-4)^2/16-(y-1)^2/33-1)((x+3)^2+(y-1)^2-0.03)((x-11)^2+(y-1)^2-0.03)((x-4)^2+(y-1)^2-0.03)(x^2+(y-1)^2-0.03)((x-8)^2+(y-1)^2-0.03)=0 [-6, 14, -4.5, 5.5]}