hyperbola question help please.

Question

tylerd · Answer

center at $$(\frac{ 117 }{ 4 }, -2.5)$$

has a vertical transverse axis of length 6.

asymptotes of y=2/9x-9 and y=-2/9x+4

tylerd · Answer

eccentricity is sqrt(85)/2

tylerd · Answer

I need help finding the focal length

tylerd · Answer

I thought it would be 2sqrt(85) but its not apparently...?

tylerd · Answer

@aum dude i need ure help

aum · Answer

For a hyperbola whose transverse axis is vertical, the slope of the asymptotes = a/b = 2/9

length of transverse axis = 2a = 6

Also c^2 = a^2 + b^2

Find a, b and c first.

tylerd · Answer

so a=3
3/b=2/9

b=27/2?

aum · Answer

Yes.  Find c which is the distance between the center and the focus.

tylerd · Answer

so it would be 2c?

tylerd · Answer

approx - 2(13.83)

aum · Answer

about 27.7.  Is that the correct answer?

tylerd · Answer

hell it didnt work....

tylerd · Answer

wtf

aum · Answer

Do they tell you how many decimal places they want the answer in?

tylerd · Answer

i did it exactly as the calc said

tylerd · Answer

but the thing is

tylerd · Answer

for eccentricity i got sqrt(85)/2 and it said it was right

tylerd · Answer

and e=c/a right?

aum · Answer

yes.

aum · Answer

Oh, you did twice the focal length. It should have been just 13.83 or 13.8 not twice that.

tylerd · Answer

so just 13.82931669?

tylerd · Answer

WORKED

aum · Answer

IDK how many decimal places they are expecting but if it is two decimal places it should be 13.83.

tylerd · Answer

it doesnt say so i just put all of them in to be safe

aum · Answer

cool.

tylerd · Answer

been on here for an hour and no one was able to help

aum · Answer

glad to be able to help.

aum · Answer

Not sure what is going on here.  I checked the text and for hyperbola  it says the focal length is 2c.  Not sure why it is not accepting 2c but accepting c.