Question

Could you please help me find the center, vertices, foci, eccentricity, retricemptotes, length of transverse axis and the length of the conjugate axis for \[\frac{ (x-4)^2 }{ 36 }-\frac{ (y-2)^2 }{ 9 }=1\]

Answer 1

asymptotes*

Answer 2

retricesymptotes LOL

Answer 3

hahaha im tired and i am trying to finish 2 last questions for my study guide for the final in like 2 days o.o

Answer 4

The center is found from the values (x-h)^2/a^2 - (y-k)^2/b^2 = 1
(h,k) --> center

Answer 5

would the center be (4,2)?

Answer 6

Eccentricity = c/a
Yes that is the center

Answer 7

Center: (4,2)
eccentricity= c/a
For the eccentricity you'll have to figure out what c is by plugging the values for a and be into the equation c^2=a^2-b^2

Answer 8

Find C, from the equation
a^2 = b^2 + c^2
solve for c.
a^2 = 36, b ^2 = 9

Answer 9

The length of the transverse axis is 2a, and the length of the conjugate axis is 2b.

Answer 10

For the foci and vertices, I'm clueless atm about..

Answer 11

okay for c^2 i got 27, so for the eccentricity it would be 27/36 which is .75?

Answer 12

e = c /a
c^2 = 27
c = 3√3

Answer 13

for the lengths would it be 2(6) and 2(3)?

Answer 14

a^2 = 36
a = 6
e = c/a
e = 3√3/6
e = √3/2

Answer 15

Yes.
12 , and 6. for the tranverse/conjugate axis

Answer 16

where did you get 12 and 6 from?

Answer 17

wouldn't it be 6 and 3?

Answer 18

2a , and 2b
i just multiplied them lol.

Answer 19

OH hahaha

Answer 20

thank you so much!!!

Answer 21

do you know how to find the asymptotes?

Answer 22

Nope.. I don't know how to find the foci/vertices either.

Answer 23

its okay, thank you so much!! :)

Answer 24

The foci are located at (4 (+/-) 3*sqrt(3), 2)