Question

FAN AND METAL
Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7).

Answer 1

any ideas how to proceed?

Answer 2

no idk how to do it can you explain

Answer 3

your material should give you a standard general setup for the equation that we can start to fill in. we should start with that.

Answer 4

we will want to find a center point, and distances from that center to a vertex, a focus, and the rest is pythag thrm

Answer 5

so how would i start

Answer 6

your material should give you a standard general setup for the equation that we can start to fill in. we should start with that.

Answer 7

if you cant find it in your lesson material, the internet has a plethera of them laying about

Answer 8

(x-h)^2/a^2 -(y-k)^2/b2

Answer 9

thats what it gives me

Answer 10

that one will do fine.

now the point (h,k) defines the center point. that will be the midpoint between vertexes, or the midpoint between focuses

Answer 11

and for a little correction, we need the equation to equal 1

(x-h)^2/a^2 -(y-k)^2/b2 = 1

can you find the center point for us?

Answer 12

how do i do that

Answer 13

the center point is an average of the extremes:

lets use the verts: (0,3) and (0,-3) now if its not obvious that (0,0) is the center between them we can go about it as:

0+0 3-3
---- , --- = (0,0)
2 2

Answer 14

adding the component parts, and dividing by 2

Answer 15

okay so the center is (0,0)

Answer 16

yes, which tells us (h,k) = (0,0), so lets use that in the setup

x^2/a^2 - y^2/b^2 = 1

now a hyperbola, is not necessarily x-y, but maybe y-x so we need to determine the proper sign placement

we are told that when x=0, y=7 by the vertex, agreed?

using that we get:

0/a^2 - 7^2/b^2 = 1; now -7^2/b^2 can never equal +1 can it? so we must change the order of the setup to:

y^2/b^2 - x^2/a^2 = 1

does this make sense?

Answer 17

i should have used the verts on that, but at my age i tend to get things mixed about ....

Answer 18

not really

Answer 19

what about it is not making sense?

Answer 20

does it make sense up to a point? or is it all just muddled?

Answer 21

do we agree that h=0, and k=0?

Answer 22

yea i get the center

Answer 23

\[\pm\frac{(x-0)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1\]
\[\pm\frac{x^2}{a^2}\mp\frac{y^2}{b^2}=1\]
we are given that the point (0,3) has to be a solution to the equation, we can use that to determine 2 things, sign usage and one of the values for a^2 or b^2
\[\mp\frac{3^2}{b^2}=1\]since a negative is never equal to a postive, that means the sign is a +, and the only way this can be equal to 1 is if b^2 =3^2
\[-\frac{x^2}{a^2}+\frac{y^2}{3^2}=1\]or written a more standard way
\[\frac{y^2}{3^2}-\frac{x^2}{a^2}=1\]
and we have only one more thing to determine to finish it up

Answer 24

in a hyperbola, the pythagorean thrm comes into play; if we call f the distance from the center to a focus point: then a^2 + b^2 = f^2 and we can solve for a^2

a^2 = f^2 - b^2

a^2 = 7^2 - 3^2