Question

A hyperbola has vertices (+-5, 0) and one focus (6, 0). What is the standard form equation of the hyperbola?

I just would like some help starting....please no direct answers :)

Answer 1

Ok! So what is standard form of a hyperbola?

Answer 2

For this one, wouldn't it be \[(y-k)^2=4p(x-h)\]
because it is opens up horizontally?

Answer 3

Hm. Are you sure that's the standard equation?

Answer 4

No...actually I am not... :(

Answer 5

I think you are mistaken that is for a parabola

Answer 6

oh yeah! lol one second

Answer 7

We are looking for a hyperbola :)

Answer 8

Okay one second lol I can't recall that formula off the top of my head right this second :(

Answer 9

Do you need me to write it for you?

Answer 10

I got it lol I just had a duh moment

Answer 11

\[\frac{ (x-h)^2 }{ a^2 }+ \frac {(y-k)^2}{b^2}=1\]
right?

Answer 12

Yes.

Answer 13

So your vertex is (h, k + a)

Answer 14

Okay

Answer 15

Jk

Answer 16

It's (h + or - a , k)

Answer 17

Okie dokie

Answer 18

Do you know what a is now?

Answer 19

5?

Answer 20

Yup! Do you know what h and k are yet?

Answer 21

(0, 0)? So the center is the origin....

Answer 22

Yes! Now we just need to know what b is.

Answer 23

b is -5, correct?

Answer 24

Nope!

Answer 25

:( I'm not sure then.....

Answer 26

Dont worry! We can find b by a^2 + b^2 = c^2. c is part of the focus, (h + or - c, k)

Answer 27

Okay....

Answer 28

So lets fill in the values to find b.
a^2 + b^2 = c^2
5^2 + b^2 = 6^2

Answer 29

ah hah!

Answer 30

so b^2 is going to be 11....so \[\frac{x^2}{25} + \frac{y^2}{11}=1\] right? I'm pretty sure that is it.....

Answer 31

Yeah checks out

Answer 32

Thank you!

Answer 33

np

Answer 34

@swagmaster47 that doesn't work because it then forms an ellipse....

Answer 35

I'm so confused.....

Answer 36

I figured out the problem

Answer 37

You put a plus instead of a subtraction symbol between the two

Answer 38

it's supposed to be plus.....but we solved for b incorrectly.

Answer 39

it was supposed to be a^2=b^2+c^2....

Answer 40

No we did b right

Answer 41

so then 11 would be -11......I think it works either way...

Answer 42

Cant be -11

Answer 43

You squared b

Answer 44

squared b? when did we square it?

Answer 45

because it just cancels out when you square a square root.......so -11 would just be plugged in for b^2.....

Answer 46

I am so lost :(

Answer 47

it can't be a^2+b^2=c^2 because that is the pythagorean theorem..

Answer 48

Technically b is sqrt(11)

Answer 49

that is true...

Answer 50

"The vertices are a units from the center, and the foci are c units from the center, with c^2 = a^2 + b^2.
Notice that in the equations for a hyperbola, unlike an ellipse, a^2 is not necessarily larger than b^2"

Answer 51

I'm so confused :(

Answer 52

https://www.desmos.com/calculator/wd0kkajkqu

Answer 53

You got it, but you put a plus between your fractions instead of a minus