Question

find the vertices, foci, eccentricity, asymptotes, length of transverse axis, length of conjugate axis for (x-4)^2/36-(y-2)/9=1

Answer 1

@cmorr23

Answer 2

Alright find the center.

Answer 3

(4,2)

Answer 4

\[\frac{ (x-4)^2 }{ 36 }-\frac{ (y-2) }{ 9 }=1\]

Answer 5

(y-2)^2

Answer 6

Ok is it vertical or horizontal?

Answer 7

Do you need help with that part?

Answer 8

horizontal

Answer 9

Yup good job. Ok so now where are your vertices?

Answer 10

Remember they're going to be to the left and right of the center.

Answer 11

So the y axis won't change.

Answer 12

it is to the left and right but i dont by how much

Answer 13

Ok so now to find the vertices you know how there are different numerators with the y in one and x in the other? You need to find the square root of the one with the x in the numerator since it is a horizontal equation. So find the square root of 36.

Answer 14

6

Answer 15

Good so the vertices are going to be 6 to the right and left of the center.

Answer 16

So now what are your vertices?

Answer 17

(10,2) and (-2,2)

Answer 18

Yup good job. So now we'll find the foci which are represented by the equation of c^2=A^2-B^2. So which number is a^2 and which one is b squared?

Answer 19

a^2= is 9 and B^2=36

Answer 20

Oops my bad I meant c^2=A^2+B^2 because it's a hyperbola. But no it's the other way around Because A squared is always the bigger number and b squared is smaller. So A squared is equal to 36 and b squared is equal to 9. So now plug those into the equation and what do you get?

Answer 21

45 and i think its square rooted

Answer 22

Yeah so your foci are plus or minus the sqrt of 45.

Answer 23

So do you know how to find eccentricity?

Answer 24

isnt it c over a?

Answer 25

Yup

Answer 26

So what's the eccentricity?

Answer 27

sqaure root of 45 and then divide it by 6.

Answer 28

i got 1.11803

Answer 29

Yup and remember the eccentricity of a hyperbola is always greater than 1 so you can check your answer that way.

Answer 30

Ok now for the the asymptotes what do you know about those?

Answer 31

Lets do the traverse axis and conjugate axis they're really easy.

Answer 32

ok show me how to find them

Answer 33

Aight so for the traverse axis find the square root of A^2 and multiply it by 2 and you're done.

Answer 34

And the same thing for the conjugate axis except you use b^2

Answer 35

So what are your axes?

Answer 36

so the transverse axis is 12 and the conjugate axis is 6

Answer 37

Yup

Answer 38

Now for the asymptotes since your hyperbola isn't in the center you've gotta subract the y coordinates and x coordinates from b/a so it'll look like this. One sec.

Answer 39

\[y +2=\pm3/4(x+4)\]

Answer 40

\[y +2=\pm3/6(x+4)\]

Answer 41

Once you work it out you get x-2y=0 and x+2y-8=0.

Answer 42

so then what?

Answer 43

Then you have to put them in slope intercept form and all that crap.

Answer 44

wow so this is the long annoying part huh

Answer 45

Yup I hate asymptotes. lol

Answer 46

slope intercept form is the y=mx+b thing right?

Answer 47

Yup THAT thing

Answer 48

One sec i'll help with the asymptotes thing I just need to check over it again.

Answer 49

ok

Answer 50

I got it this time I wrote the equation wrong so the formula for asymptotes is y-k=b/a(x-h)
With K=2 h=4 b=3 and a=6.

Answer 51

So then you plug it in and it looks like this y-2=3/6(x-4).

Answer 52

Next you can simplify 3/4 to being 1/2 then you multiple 1/2 times x-4 and you get y-2=1/2x-2

Answer 53

Then you try to get the y by itself and add two to each side giving you y=1/2x because the 2's cancel each other out.

Answer 54

Then to get your other asymptote you do the exact same thing except instead of having regular b/a you have negative b/a.

Answer 55

Got it? Sorry I took so long i've been working all day.

Answer 56

its cool thank you so much for all of the help

Answer 57

Yup your welcome.

Answer 58

do you know a website that could help me graph the two equations that i asked you to help me on?

Answer 59

are you still there?

Answer 60

Geogebra Is a good one that's what I used for my class it's not a website though it's a program.

Answer 61

I think there's actually an online version of it too that you don't have to download. But I've only used the one you download it's free by the way.