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#?

1 Answer
Massimiliano
Feb 18, 2015

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#

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