Question

Help with application of ellipses and hyperbolas

Answer 1

Find the equations of the asymptotes of the hyperbola described by the equation:

Answer 2

\[\frac{ x^2 }{ 36 }-\frac{ (y+6)^2 }{ 11 }\]

Answer 3

To find the asymptotes it is represented by the equation \[y=k+/-b/a(x-h)\]

Answer 4

@freckles question what squares into 11

Answer 5

also i forgot to add the = 1 to the equation

Answer 6

\[\frac{x^2}{6^2}-\frac{(y+6)^2}{(\sqrt{11})^2}=1 \\ \sqrt{11} \approx 3.317\]

Answer 7

so just the sqrt 11 mang that was simple

Answer 8

yah

Answer 9

you could say a is sqrt(36) if you wanted

Answer 10

\[k=6+/-\frac{ \sqrt11 }{ 6 }(x-0)\]

Answer 11

k is -6

Answer 12

oh and k is y lol

Answer 13

I'm sorry let me say that a different way
your 6 is -6 since k is -6

the other thing I was trying to say is your named the equation k instead of y

Answer 14

wowsers. the answer choices are this though which is why i am confused.

Answer 15

I guess you could call it k
just don't give the name of the equation confused with what I was trying to tell you about your other k

Answer 16

ohh so y=k and k =-6 or y = -6

Answer 17

k is -6
y is not really k
y was the name of the equation before you or the software you are using named it k

Answer 18

ohhh ok

Answer 19

so the equation full and simplified would by y=-6+/- sqrt11/6x?

Answer 20

\[y=k \pm \frac{b}{a}(x-h) \\ k=-6 \\ b=\sqrt{11} \\ a=\sqrt{36}=6 \\ h=0 \\ y=-6 \pm \frac{\sqrt{11}}{6}(x-0) \\ y=-6 \pm \frac{\sqrt{11}}{6}x \]
and the way I'm fixing to use k is totally different than the way I used it above
they renamed your equation k instead of y
this k is not the k from earlier
\[k=-6 \pm \frac{\sqrt{11}}{6} x\]

Answer 21

Alright. Thank you for explaining it :D

Answer 22

I think they meant y though

Answer 23

Same I substituted for y in my first equation so i could get k to the other side
which was indeed wrong

Answer 24

names can get confusing when you are using them for two different people

Answer 25

If you call two guys in a room Ted and then call on Ted to answer a question, the Teds can be confused on to whom you are speaking.

Answer 26

loool great analogy 10/10

Answer 27

But it doesn't happen in math a lot.

Answer 28

indeedly 2 teds very rare

Answer 29

lol I never known a ted

Answer 30

same lol

Answer 31

even 1 is rare :p

Answer 32

well john probably would have been better

Answer 33

Hmm or will

Answer 34

yah