Question

Hyperbola question, check my work?

Answer 1

The hyperbola I'm given:
(x-4)^2/36 - (y-2)^2/9 = 1

Center: (4, 2)
Vertices: (-2, 2) and (10, 2)
Foci: (11, 2) and (-3, 2)
Eccentricity: 3sqrt3 (approx. 5.2)
Asymptotes: y = 1/2x and y = -1/2x + 4
Length of transverse axis: 12
Length of conjugate axis: 6

Answer 2

I graphed the hyperbola and found all the info (center, foci, etc.). Is it all correct?

Answer 3

I don't agree with the part
`Foci: (11, 2) and (-3, 2)`
`Eccentricity: 3sqrt3 (approx. 5.2)`
other than that, it looks good

Answer 4

How do these look? I redid them and got:
Eccentricity: sqrt(3)/12
Foci: (10 - 3sqrt3, 2) and (-2 - 3sqrt3, 2)

Answer 5

use the formula
\[\Large E = \sqrt{1+\left(\frac{b}{a}\right)^2}\]
E = eccentricity of hyperbola
a^2 = 36 ---> a = 6
b^2 = 9 ----> b = 3

Answer 6

for the foci, use this
http://www.mathwords.com/f/foci_hyperbola.htm

Answer 7

I thought the formula for the eccentricity was e = c/a?

Thanks for the link! I'll go check it out.

Answer 8

`I thought the formula for the eccentricity was e = c/a?`
which source is saying that?

Answer 9

oh I see what you mean

Answer 10

\[\Large E = \frac{c}{a}\]
\[\Large E = \frac{\sqrt{a^2+b^2}}{a}\]
\[\Large E = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2}}\]
\[\Large E = \sqrt{\frac{a^2+b^2}{a^2}}\]
\[\Large E = \sqrt{\frac{a^2}{a^2}+\frac{b^2}{a^2}}\]
\[\Large E = \sqrt{1+\frac{b^2}{a^2}}\]

Answer 11

Haha. Yeah. So either way for the eccentricity?

Answer 12

http://www.mathsisfun.com/geometry/eccentricity.html

Answer 13

yeah either way is fine

Answer 14

you should get approximately E = 1.11803 for the eccentricity

Answer 15

c^2 = b^2 - a^2 right?

Answer 16

the once you know E, you can use it to find c

E = c/a
c = E*a

to help find the two focus points
(h-c, k)
(h+c, k)

Answer 17

a^2+b^2 = c^2

Answer 18

36 + 9 = c^2
45 = c^2
so c = sqrt 45
Then e = sqrt45/36 which is 0.186
What am I doing wrong...

Answer 19

a^2 = 36
a = 6

For e = c/a, you plugged in the wrong 'a' value

Answer 20

Oh! Okay.

Now for the foci... what is the fomula?

Answer 21

foci
(h-c, k)
(h+c, k)

where
(h,k) is the center

Answer 22

this only works if the hyperbola opens left and right

Answer 23

(4 + 1.118, 2) and (4 - 1.118, 2)
I think I got it.

Answer 24

If it opened up/down, would the foci be
(h, k - c)
(h, k + c)

Answer 25

\[\Large \left(4-3\sqrt{5}, 2\right) \ \ \text{and} \ \ \left(4+3\sqrt{5}, 2\right)\]
if you want to state the foci exactly

Answer 26

```
If it opened up/down, would the foci be
(h, k - c)
(h, k + c)
```
yes that is correct

Answer 27

btw the foci aren't (4 + 1.118, 2) and (4 - 1.118, 2)

Answer 28

c is not equal to 1.118

Answer 29

Thank you SO much for you help

Answer 30

you're mixing up the eccentricity and c

Answer 31

Oohh... okay

Answer 32

Got it, thanks :)

Answer 33

no problem