Question

Find the standard form of the equation of the hyperbola with the given characteristics.
vertices: (2, - 6), (2, - 2)
foci: (2, - 8), (2, 0)
I know this much so far:
(y+4)^2 (x-2)^2
------ - ------
? ?

Answer 1

the y is the heavy side; so this opens up and down

Answer 2

teh distance between the foci is 8; so each foci is 4 from center; that 4 is either a or c :)

Answer 3

ok so a

Answer 4

we have 4 between the vertices; so each vertex is 2 from center

Answer 5

right how do I find b^2 then?

Answer 6

under the y we have 2^2 right? let me look this up some more :)

Answer 7

ok I think actually I have to know what c^2 is
cause C^2=a^2+b^2

Answer 8

I'm thinkin' b^2=16...?

Answer 9

tell me your understanding of a b and c; what do they measure?

Answer 10

the foci are c units from the center
the vertices are a units from the center

Answer 11

so; a^2 + b^2 = c^2 then right?

Answer 12

2 + b^2 = 64

b^2 = 62 then

Answer 13

I think so....but that doesn't eem right maybe it's minus?

Answer 14

my options for b^2 are 12 or 16 here but I don't see how they get either of those

Answer 15

like this right?

Answer 16

i used 8 instead of 4 :)

Answer 17

i think so...

Answer 18

16-4 = 12

Answer 19

b^2 = 12

Answer 20

oh ok I was gonna say 16 cause 4^2=16 but 12 works too

Answer 21

the minor axis is always smaller in number than the major axis

Answer 22

since the major is the y on this one; and it is 16; the other has to be less

Answer 23

or maybe i messed that up in thougth

Answer 24

ugh... idk this is all a review for my final and I am thinking that so far i'm not doing too hot. YIKES

Answer 25

y^2 x^2
--- - --- = 1
4 12

b^2 = focal distance(c)^2 - vertex distance(a)^2

in this case; the hyperbola; c is the greatest distance beacusae it is used as the hypotenust

in the ellipse; a is the greatest distance becasue IT is used for ITS hypotenuse