Question

Another hyperbola question below :)

Answer 1

26. Find the equation of the hyperbola.
Vertices at (-4, 0) and (4, 0); asymptote at the line y = 2x

Used y - k = \(\pm \dfrac ba\)(x-h)
Center is (0, 0) because y - k = y, and x - h = x
a = 4 because of the vertices

y = \(\pm\dfrac b4\)(x)
b = 8

\(\dfrac {x^2}{ a^2} - \dfrac{y^2}{b^2}=1\)
b^2 = c^2 - a^2
64 = c^2 - 16
80 = c^2
\(\dfrac{x^2}{16} - {y^2}{80} = 1\)

Answer 2

\(\dfrac{x^2}{16} - \dfrac {y^2}{80} = 1\)

Answer 3

you got b=8, correct
b^2 = 64

In the equation of hyperbola, you don't need 'c'

Answer 4

-y^2/b^2 = - y^2/64 ....

Answer 5

ooohhhh... now I feel dumb...

Answer 6

x^2/16 -y^2/64 = 1
would be correct.