Question

Question #083fa

Answer 1

Center: `(3/8,13/2)`
Focal length: `10`
eccentricity: `5/4`

Explanation:

Here is a reference for hyperbolas; the formulas and facts that I will say that we know can be found in this reference.

The center is at the intersection of the two asymptotes:

`y=(4/3)x+6`
`y=-(4/3)x+7`

`0 = -8/3x+1`

`8/3x=1`

`x = 3/8 larr` the x coordinate of center

Substitute into the first equation:

`y = (4/3)(3/8)+6`

`y = 1/2+6`

`y = 13/2 larr` the y coordinate of the center

Substitute into the standard Cartesian form for a hyperbola with a vertical transverse axis:

`(y-13/2)^2/a^2-(x-3/8)^2/b^2=1`

We know that the length of the vertical axis is `2a`

Given: `2a = 16`

`a = 8`

`(y-13/2)^2/8^2-(x-3/8)^2/b^2=1`

From the slopes of the asymptotes, we know that:

`a/b = 4/3`

Substitute 8 for a:

`8/b = 4/3`

`b = 6`

`(y-13/2)^2/8^2-(x-3/8)^2/6^2=1`

The focal length, c, is:

`c = sqrt(a^2+b^2)`

`c = sqrt(8^2+6^2)`

`c = 10`

We know that the eccentricity is

`c/a`

`10/8`

`5/4`