Write the standard form of the hyperbola that has a center of (0, 0), vertices at (3, 0) and (-3, 0), and slopes of the asymptotes are 4/3 and -4/3.

Question

anonymous · Answer

see ill tell u the method

anonymous · Answer

the hyperbola is of the form x^2/a^2 + y^2/b^2 = 1

anonymous · Answer

a = 3 as the vertices are at 3,0 and -3,0

anonymous · Answer

if the angle bw asymptotes is t, the eccentricity e = sec (2t)

anonymous · Answer

and we know 
b^2/a^2  = e^2   - 1

anonymous · Answer

so find b frm this..get it?

anonymous · Answer

b is 72/25

anonymous · Answer

get it??????

amistre64 · Answer

that part tends to get me; why would 'b' not be 4?

anonymous · Answer

yeah i got it i think

anonymous · Answer

what is the actual equation though

amistre64 · Answer

x^2    y^2
--- - ---- = 1  right?
  9      b^2

anonymous · Answer

so there is no b^2?

amistre64 · Answer

im thinking b^2 = 16 .... but i could be wrong

amistre64 · Answer

hims smart about this stuff which makes me doubt my b^2 :)

anonymous · Answer

yeah ur right amistre...

amistre64 · Answer

yay!!  im right

anonymous · Answer

im lovesick 2 amistre..bound to go wrong smwhere..lol

amistre64 · Answer

lol   trees pic is cute...

anonymous · Answer

y du think i jumped onto her question? dont mind tree..compliments

anonymous · Answer

:)

amistre64 · Answer

i thought maybe to give her an answer :)

anonymous · Answer

yes..bt with ulterior motives 2..lol

anonymous · Answer

lol :))))))))))))))))

anonymous · Answer

never knew beautiful people could give parentheses life with their smiles..;)

anonymous · Answer

lol thanks