Question

I need help with hyperbolas please? Posting the hyperbola equations and questions in the comments.

Answer 1

x^2 - y^2 = 1

Answer 2

\[\frac{ (x-1)^{2} }{ 2 }-\frac{ (y+3)^{2} }{ 2 }=1\]

Answer 3

The center of the hyperbola is (1, -3), right?

Answer 4

yes

Answer 5

Also, I need to find the lengths of the transverse and conjugate axes, the slopes of the asymptotes, and the foci, but I'm really not quite sure how to do that...

Answer 6

the slope of the asymptotes is the easiest part
if you have
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] the slopes are \(m=\pm\frac{a}{b}\) in this case \(\pm1\)

Answer 7

Oh my, I wrote the equation wrong, a^2 should be 64 and b^2 should be 81.

Answer 8

So, then the slopes of the asymptotes will be 64/81? or would that be vice versa, 81/64? is a always greater than b?

Answer 9

ok no problem
in this case you have
\[\frac{(x-1)^2}{8^2}-\frac{(y+3)^2}{9^2}=1\]

Answer 10

the slope is not \(\pm\frac{a^2}{b^2}\) it is \(\pm\frac{a}{b}\)

Answer 11

in your case \(\pm\frac{8}{9}\)

Answer 12

does that need to be simplified any further or can it stay as a fraction?

Answer 13

actually it is wrong

Answer 14

should be \(\pm\frac{9}{8}\) i had it upside down

Answer 15

oh, okey dokey. so does that need to be simplified or can it stay as a fraction?

Answer 16

not sure what you mean by "simplify"
there is no common factors, so you cannot reduce it
just leave it as it is

Answer 17

I mean't, like, turn it into a decimal, but I figured it should just be left as is, so thank you (:

Answer 18

Is there a formula I use to find the foci and the lengths of the axes?

Answer 19

i think the lengths are just 8 and 9

Answer 20

that is, semi major is 8
semi minor is 9

not much to that part

Answer 21

finding the foci starts with having an idea of what this looks like

Answer 22

there is my lousy picture, but i drew it to show that the foci are to the right and left of the center, which is \((1,-3)\)

Answer 23

Yes, I know where they're located, I'm just not sure how to find their coordinates using the equation.

Answer 24

since \(a^2+b^2=64+81=145\) you know \(c=\sqrt{145}\)

Answer 25

and since the center is \((1,-3)\) the foci are at \((1-\sqrt{145},-3)\) and \((1+\sqrt{145},-3)\)

Answer 26

Oh, okay, so the square root of 145 is 12.04159458, so that means the foci are (-11.04159458, -3) and (13.04159458, -3), right?

Answer 27

Is it okay to round those numbers? Will it still be accurate enough?

Answer 28

wait a second, is this hyperbola vertical? i remember reading something a while back saying that a (the major axis) is always longer (greater) than b (the minor axis), and if they're flipped, wouldn't that make it vertical?

Answer 29

yeah i drew it wrong
answer was correct though
you can check it here
http://www.wolframalpha.com/input/?i= \frac{%28x-1%29^2}{8^2}-\frac{%28y%2B3%29^2}{9^2}%3D1

Answer 30

Okay! Thank you!