Question

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)?

Answer 1

`(x-4)^2/16-(y-1)^2/33=1`

Explanation:

As the two vertex are `(0,1)` and `(8,1)` and focus is `(-3,1)`

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

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

`(x-4)^2/16-(y-1)^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]}