anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
see ill tell u the method
anonymous
  • anonymous
the hyperbola is of the form x^2/a^2 + y^2/b^2 = 1
anonymous
  • anonymous
a = 3 as the vertices are at 3,0 and -3,0

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More answers

anonymous
  • anonymous
if the angle bw asymptotes is t, the eccentricity e = sec (2t)
anonymous
  • anonymous
and we know b^2/a^2 = e^2 - 1
anonymous
  • anonymous
so find b frm this..get it?
anonymous
  • anonymous
b is 72/25
anonymous
  • anonymous
get it??????
amistre64
  • amistre64
that part tends to get me; why would 'b' not be 4?
anonymous
  • anonymous
yeah i got it i think
anonymous
  • anonymous
what is the actual equation though
amistre64
  • amistre64
x^2 y^2 --- - ---- = 1 right? 9 b^2
anonymous
  • anonymous
so there is no b^2?
amistre64
  • amistre64
im thinking b^2 = 16 .... but i could be wrong
amistre64
  • amistre64
hims smart about this stuff which makes me doubt my b^2 :)
anonymous
  • anonymous
yeah ur right amistre...
amistre64
  • amistre64
yay!! im right
anonymous
  • anonymous
im lovesick 2 amistre..bound to go wrong smwhere..lol
amistre64
  • amistre64
lol trees pic is cute...
anonymous
  • anonymous
y du think i jumped onto her question? dont mind tree..compliments
anonymous
  • anonymous
:)
amistre64
  • amistre64
i thought maybe to give her an answer :)
anonymous
  • anonymous
yes..bt with ulterior motives 2..lol
anonymous
  • anonymous
lol :))))))))))))))))
anonymous
  • anonymous
never knew beautiful people could give parentheses life with their smiles..;)
anonymous
  • anonymous
lol thanks

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