Question

How can I graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.

Answer 1

ok, let's take a peek

Answer 2

vertices are at
(0, 8)
(0, -8)

notice "x" doesn't change, "y" does

so is a vertical traverse axis
so is going up and down

Answer 3

the distance from the vertices to the (h,k) center, is "a"
so, one vertex plus the other, half that, is where the center is
in this case is moving up 8, down 8
so the center is clearly (0, 0)

Answer 4

It's a parabola that opens up and down right? :D

Answer 5

hehe

Answer 6

parabolas only open either down or up, not both :)

Answer 7

Alright, so the center would be 0,0? What equation would I use to graph that? Like how do I set up the equation

Answer 8

well, I assume you know what the equation of a hyperbola is?

Answer 9

It'd be two parabolas then? :D

Answer 10

well, yes, a hyperbola is pretty much 2 parabolas, mirroring each other

Answer 11

you do know what a hyperbola is?

Answer 12

as in, the exercise wouldn't be here if you hadn't covered that already

Answer 13

and I also assume you know a hyperbola equation

Answer 14

Yeah, that's what we're studying right now.
I know how to graph them when I have the equation just I'm not sure how I get the equation when I only have the vertices.

Answer 15

ok, so your (h, k) center is a (0, 0)
from there it moves up 8
down 8
to the vertices

the distance from the center to either vertex is "a"

Answer 16

so a = 8

Answer 17

now, your foci are at
(0, 10) (0, -10)

so is going up 10
and down 10

the distance from the center to either focus is "c"
so
c = 10

now
\(c^2 = a^2 + b^2 \) as in pythagorean theorem
so \(b^2 = c^2 - a^2\)

solve that for \(b^2\)
a = 8, so \(a^2 = 64\)

Answer 18

and then put the values in the hyperbola equation :)

Answer 19

so, I solve \[10^2 = 8^2 + x^2\]

Which eould be 2^2?

Answer 20

it might help to keep your a and b parts under the x and y parts

b is conventionally an up down measure, and a is left to right measure.

using slope as y/x then b/a conforms nicely to your under xy parts

Answer 21

Now I'm just confused. :/

Answer 22

amistre64 is correct, which one goes under which one?

well "a" always goes to the left-hand-side fraction and "b" to the other :)

and since it's a vertical traverse axis hyperbola, up and down
the fraction with the POSITIVE sign will the one with the "y"

Answer 23

well, solve for \(x^2\)

Answer 24

so, what does \(10^2 = 8^2 + b^2\) give you for \(b^2\)

Answer 25

there are 2 parts to an ellips or a hyperbola\[\frac{x^2}{a^2}~and~\frac{y^2}{b^2}~=~1\]
since your verts are at (0,\(\pm\)y): (0,\(\pm\)8)
\[\frac{0^2}{a^2}~+~\frac{8^2}{8^2}~=~1\]
\[-\frac{0^2}{a^2}~+~\frac{8^2}{8^2}~=~1\]rewriting for asthetics
\[\frac{y^2}{8^2}-\frac {x^2}{a^2}=~1\]solving a has already been addressed

Answer 26

a^2 would be 36 so a = 6?

Answer 27

well, a = 8, \(8^2 \ne 36\)

Answer 28

so, what did you get for \(b^2\)?

Answer 29

oh, is \[b^2 = 36\]

Answer 30

ohh.... I guess you meant b = 6

yes

Answer 31

right, so, keep in mind that is going vertical, so, in a hyperbola you have 2 fractions, one is negative, one is positive

the hyperbola goes on the direction of the POSITIVE variable
if it's going upwards, that means "y" fraction is POSITIVE

Answer 32

yes, the "missing" value is 6.

Answer 33

so, how does your hyperbola look like? :)

Answer 34

I guess I was looking for the equation, but yes, that looks right :)