Question

Help with hyperbolas? :D

Answer 1

Find the vertices and foci of the hyperbola with equation:

Answer 2

The answer choices don't match my answers!? @mathstudent55 @satellite73 @surjithayer

Answer 3

Can you show me your steps of how you're solving it? Then I can show you where you need to fix it. =)

Answer 4

I'm a little confused though, this isn't an equation.

Answer 5

I said it was parallel to the x axis because the x^2 was positive, so the center=(5,1). a=9, b=12, so c=15. But I got (15, 0), and (-15, 0) for the foci, and (12,0), and (-12, 0) for the vertices.

Answer 6

What???

Answer 7

It said online to use c for the foci and a for the vertices. ?

Answer 8

does this help
http://www.purplemath.com/modules/hyperbola2.htm
?

Answer 9

No.. Actually, that is the exact page I looked at. They used c for the foci and a for the vertices.. I tried that..

Answer 10

then try the opposite it'll work

Answer 11

What do you mean? -_-

Answer 12

you are close
when you did this step
a=9, b=12, so c=15.
that is good

but not this part:
But I got (15, 0), and (-15, 0) for the foci, and (12,0), and (-12, 0) for the vertices.

you find the vertices and the focus starting from *the center* (5,1) (not the origin)

Answer 13

How do I find the center again?

Answer 14

BG: There's a lot of good info at this URL:

http://www.purplemath.com/modules/hyperbola.htm

The general form of the equation of a hyperbola is \[\frac{ (x-h)^2 }{ a^2 }-\frac{ (y-k)^2 }{ b^2 }=1\] Please note that there must be a " 1 " on the right side.

Notice those characters h and k? They are the coordinates of the CENTER of your hyperbola. Could you please now find the center of this hyperbola?

Answer 15

the center are the numbers from (x-h) and (y-k)
(h,k) is the center. Here the center is 5,1

Answer 16

Oh, okay: 5 and 1. Got it. I knew that, I just forgot. Depressing. So I go from there to??

Answer 17

Next important fact: Determine which variable appears first, x or y. If x appears first, then your hyperbola is a horizontal one. If y first, a vertical one.

So here, your parabola is

veritical?
horizontal?

Answer 18

now do what you did before. keep the same x value as the center, but add or subtract "a" to get to the vertex, add /subtract c to get to the focus

Answer 19

** keep the same y value as the center

Answer 20

Horizontal. :3

Answer 21

Phi: wouldn't " a " be the distance from the center to the vertex (not the focus)?

Answer 22

Horizontal. :3. Cool.

Answer 23

So... What? (1, 9)?

Answer 24

you want to stay at the same y value as the center (5,1)
and move sideways

Answer 25

Your a^2 is 81, so your " a " is 9.

Using that info, are you able to locate the two vertices now?

Answer 26

So I keep the 1. But my vertices can't be 10 and -10..

Answer 27

So, BG: Taking the center of your hyperbola to be (5,1), add a = 9 to 5, and then subtract a = 9 from 5. What two points result? These points represent your vertices.

Answer 28

the center x value is 5. you move +9 or -9 from there to get to the vertex's x

Answer 29

But 5 is the x value, you said y..

Answer 30

yes, the vertex keeps the y value of 1, but its x value is 9 away from 5

Answer 31

As phi has implied, the two vertices and the center are on the same, horizontal line. You'll soon see that the foci are also on this horizontal line, y=1.

Answer 32

Oh, okay. So our vertices are (14, 1), and (-4,1)?

Answer 33

"9 away from 5" implies that you should first add 9 to 5, and then subtract 9 from 5. What are the resultant x-coordinates? The y-coordinate stays 1.

Answer 34

That's great! Perfect. Congrats.
Now, what is your value for b? Look carefully at the given equation, and then look at the general equation. Identify b^2 by comparison, and then identify b.

Answer 35

Got that..

Answer 36

Our dog just ran away. Be Right Back!!!

Answer 37

catch that dog

Answer 38

dont let hin out the gate

Answer 39

get him now

Answer 40

We got him. Gosh, I really don't like that dog..

Answer 41

b=12 @mathmale

Answer 42

you use c to find the focus

Answer 43

So, c=15. So this time, I leave the x value the same, and add/ subtract to/from the y?

Answer 44

no, it's the same as the vertex. the focus and vertex are on the same y=1 line

Answer 45

So I would add 15 to 5, and subtract? So my foci are (-10,1) and (20,1)? :3

Answer 46

yes, here is a graph (using geogebra)

Answer 47

Awesome! Thanks so much you guys! God bless you! I'll probably be back later.. Lol.

Answer 48

Again, the illustration in

http://www.purplemath.com/modules/hyperbola2.htm

will likely help you keep " a, " " b, " and " c " straight for a hyperbola.

a=distance of vertex from center
b= length of "minor" axis
c= distance of focus from center.

Best to you, BG.

Answer 49

u ree tee hee ree