hyperbola questions giving medals for help.

Question

tylerd · Answer

with asymptotes at
$$y=\frac{ 3 }{ 2 }x+2$$
and$$y=-\frac{ 3 }{ 2 }x+4$$

so the center is at $$(\frac{ 2 }{ 3 },3 )$$

tylerd · Answer

I need help finding the focal length and eccentricity

anonymous · Answer

STRIP lol jk

tylerd · Answer

will save that for later

anonymous · Answer

i suck at math so cant help ya

tylerd · Answer

oh wait i think i know, i didnt see it says it has a vertical transverse axis...

tylerd · Answer

ok i solved the eccentricity its
$$\frac{ \sqrt{13} }{ 3 }$$

still need help finding the focal length

tylerd · Answer

i thought it would be sqrt(13) but its not accepting that as the right answer..

tylerd · Answer

welp **** guessed it and got it wrong

tylerd · Answer

gonna have to generate a new problem and start over now. But if anyone finds how to get the focal length thats all i need help with.

tylerd · Answer

have an idea i think will work