Question

How do you find the center of the hyperbola, its focal length, and its eccentricity if a hyperbola has a vertical transverse axis of length 8 and asymptotes of `y=7/2x-3` and `y=-7/2x-1`?

Answer 1

The answers are:

Center `C(2/7,-2)`, focal axis `8/7sqrt53`, eccentricity `e=sqrt53/7`.

Since this hyperbola has a vertical transverse axis (the "`-1`" at the second member) with center in `C(x_c,y_c)` (not the origin because the asymptotes don't pass from the origin), the equation is:

`(x-x_c)^2/a^2-(y-y_c)^2/b^2=-1`.

First of all let's find the center that is at a point that is on the two asymptotes. So let's solve the system:

`y=7/2x-3`
`y=-7/2x-1`

We can make a subtraction:

`y-y=7/2x-3-(-7/2x-1)rArr7/2x-3+7/2x+1=0`

`7x=2rArrx=2/7`

and:

`y=7/2*2/7-3rArry=-2`.

`C(2/7,-2)`

Since the asymptotes have slope:

`m=+-b/a`,

and the vertical axis is `2b`, than:

`b/a=7/2`

`2b=8rArrb=4`

so:

`4/a=7/2rArra=4*2/7rArra=8/7`.

The equation of the hyperbola is:

`(x-2/7)^2/(64/49)-(y+2)^2/16=-1`.

The focal axis is named `2c`, such as `c^2=a^2+b^2`, so:

`c=sqrt(64/49+16)=sqrt((64+784)/49)=sqrt(848/49)=sqrt((16*53)/49)=`

`=4/7sqrt53`, so the focal axis is `8/7sqrt53`.

The eccentricity is, in this case, is: `e=c/b=(4/7sqrt53)/4=sqrt53/7`