Question

What is the eccentricity of the graph of (x-2)^2/121 - (y+6)^2/64=1?

Answer 1

wow, it's cal2 I need time to look up at my note, friend.

Answer 2

thats alright :)

Answer 3

Is this really calc 2?

Answer 4

I thought this was like Alg II

Answer 5

Well it's suppose to be alg 2 lol

Answer 6

@Loser66 I think you're getting a little too ahead of yourself friend xD

Answer 7

@DebbieG here, Let her do it, Please @DebbieG

Answer 8

This is eccentricity of an ellipse, right?

Answer 9

I believe so

Answer 10

lol @Loser66 thanks for your faith in me but I'm not familiar with this topic! Was just looking it up....
http://www.mathsisfun.com/geometry/eccentricity.html

Answer 11

Isn't it this:

Answer 12

Ness and the asker said that it's alg and I never took alg before. I just mess it up with my cal2 hehehe.. so, I don't know how to solve @DebbieG

Answer 13

Come on loser, you know me bud ;P

Answer 14

http://www.mathwarehouse.com/ellipse/eccentricity-of-ellipse.php

Answer 15

Anyways, i gtg, tell me how it goes!

Answer 16

Thanks for the help :)

Answer 17

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

If I'm interpreting this correctly, then isn't it:

\[\Large e=\frac{ \sqrt{121+64} }{11 }=\frac{ \sqrt{185} }{11 }\]

Answer 18

hahaha, it's exactly the answer from my mess but they said that it's not cal2

Answer 19

its algebra 2 lol and thank you for helping, you guys are great :)

Answer 20

@DebbieG they said that it's an ellipse. I passed out

Answer 21

LOL well, like I said, this isn't a topic I'm very familiar with but it seems like pretty straight forward application of the formula. :)

It's a hyperbola.... Wolfram said so! ;)

Answer 22

ooooh wolfram said so? hehehe... a little bit confidence is back to me. I solved it by looking up my note, not from that tool. OOOOOK , no matter what happen. I .... give up.