Question

Find the vertices and foci of the hyperbola with equation x squared over sixteen minus y squared over forty eight = 1

Answer 1

\(\dfrac{x^{2}}{16} - \dfrac{y^{2}}{48} = 1\)

Cneter is (0,0), right?
Vertices are in the x-direction a distance of 4. Why?
Fake Vertices are in the y-direction a distance of \(4\sqrt{3}\). Why?

What is the relationship between a, b, and c for an hyperbola?

Answer 2

im sorry, this was the wrong question. could you help me with another thats similar to this?

Answer 3

@tkhunny

Answer 4

Find the vertices and foci of the hyperbola with equation quantity x plus 5 squared divided by 36 minus the quantity of y plus 1 squared divided by 64 equals 1.

Answer 5

It's almost exactly the same written in this standard form.

You tell me where the center is.

Answer 6

(5,1) ?

Answer 7

No good. (-5,-1)

Get used to it being (x-h) and (y-k)

Answer 8

oh okay. so to find the foci i use (h+c,k) right? what would c be?

Answer 9

Don't get ahead of your self. Let's use the information we have.

With +x^2 and -y^2, the vertices will be in the x0direction. Do you believe this?

Answer 10

Yes, the hyperbola will be opening left/right

Answer 11

Okay, now let's look at a. a = 6. Do you see this?

Answer 12

yes because the sq rt of 36 is 6

Answer 13

Perfect. This makes the vertices where?

Answer 14

(1,-1)

Answer 15

That's one. Where is the other? You must go both ways, left and right.

Answer 16

(-11,-1) ?

Answer 17

Awesome. There are two vertices.

Okay, now find the value of b.

Answer 18

sq rt of 64. so 8

Answer 19

Very good. This information will pin down the co-vertices. Where are they?

Answer 20

are co-vertices the sam as foci?

Answer 21

im not sure what co-vertices are

Answer 22

No. Co-vertices would be vertices if the hyperbola went the other way. I like to call them "fake vertices".

Answer 23

Center is (-5,-1)
With b = 8, co-vertices are (-5,-9) and (-5,7) - just like the vertices, but in the other direction. Make sense?

Answer 24

Oh okay, i get it

Answer 25

We're almost there. Given a and b, how do we find c?

Answer 26

a^2 + b^2 = c^2 ?

Answer 27

Perfect. Calculate c.

Answer 28

10

Answer 29

...and use that to find the foci!!

Answer 30

so it would be (-15, -1), (5, -1)??

Answer 31

Yes! Excellent. There are two more things that are often asked.

Asymptotes?
Eccentricity?

Do you need to supply these, or are we done?

Answer 32

no, those arent in the question. thank you so much! i understand it way better now.

Answer 33

Perfect. When you get to those last two, come on back and we can drag through that, too.

Answer 34

Note: The real trick, here, was just to proceed one thing at a time. There is a lot of stuff. Just plod through it and don't try to jump to the end. :-)

Answer 35

yes, it was much easier step by step