Question

Please help! Graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.

Answer 1

i assume the graphing technology needs an equation for the hyperB?

Answer 2

the general setup for a hyperB is:
\[\frac{(x-c_x)^2}{a^2}-\frac{(y-c_y)^2}{b^2}=1\]
or
\[\frac{(y-c_y)^2}{a^2}-\frac{(x-c_x)^2}{b^2}=1\]

depending on how the hyperB opens

Answer 3

yes, but i don't understand how i find those variables with what the question gives me /:

Answer 4

well, half the distance between your vertexes is "a"

the midpoint between your vertexis gives you (Cx,Cy) your centers for starters

Answer 5

by convention; "c" is the measure from center to focus; such that: a^2 + b^2 = c^2 if i recall it right

Answer 6

can you tell me what you thing the center point for this thing would be?

Answer 7

but how do i find a and b without an equation?

Answer 8

you are given the values for a and c; and use the pythag thrm to find b

Answer 9

like i would have a^2 = 144
a = 12 ... and lets say i was finding a minor axis, i would do 2a ... 2(12) = 24

Answer 10

jsut remember that "a" is the distance from center to vertex; you are given both vertex points to begin with .... right?

Answer 11

(0, 8) and (0, -8) are the verices , yes