Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/2.x.

Answer 1

For the standard form of
\(\Large \dfrac{y^2 }{a^2 } -\dfrac{x^2 }{b^2 } =1\)

the vertices are \(\Large (0, \pm a)\)

and you have vertices as (0,\(\pm\)6)

so you directly get the value of 'a' as ?

Answer 2

i'm not sure?

Answer 3

Compare
\((0,\pm a)\) with \((0,\pm 6)\)

Answer 4

im sorry, i dont understand :(

Answer 5

ok, let me give the steps, see whether you get them, if u have any doubt in any step, then ask :)

Answer 6

ok

Answer 7

\(Vertices \equiv (0,\pm a) \equiv (0,\pm 6) \\ \Large \implies a=6 \)

Asymptotes for this hyperbola are
\(\Large y = \pm \dfrac{a}{b} x\)

and we have
\(\Large y = \pm \dfrac{3}{2} x\)

so,
a/b = 3/2
and we already found a=6
so

6/b =3/2
6*2 =3*b

b= 12/3 = 4


so the equation of hyperbola is
\(\Large \dfrac{y^2 }{6^2 } -\dfrac{x^2 }{4^2}=1\)

or

\(\Large \dfrac{y^2 }{36 } -\dfrac{x^2 }{16}=1\)

thats it! :)

Answer 8

O.o that looks simple enough with the formulas. thanks

Answer 9

welcome ^_^

Answer 10

do you think you could help me with a few more?