Question

Find the vertices and foci of the hyperbola with equation (x+4)^2/9 - (x-5)^2/16 - 1

Answer 1

\[\frac{ (x+4)^{2} }{ 9 }-\frac{(x-5)^{2}}{16-1}\]
Right?

Answer 2

minus one on the whole thing

Answer 3

So...
\[\frac{ (x+4)^{2} }{ 9 }-\frac{(x-5)^{2}}{16}-1\]

Answer 4

yes

Answer 5

wait, the x and y need to be flipped

Answer 6

Wait you just need a y

Answer 7

which side

Answer 8

right

Answer 9

\[\frac{ (y+4)^{2} }{ 9 }-\frac{(x-5)^{2}}{16}-1\]

Answer 10

Oh, I swear I read left

Answer 11

\[\frac{ (x+4)^{2} }{ 9 }-\frac{(y-5)^{2}}{16}-1\]

Answer 12

\[\frac{ (x+4)^2 }{ 9 } - \frac{ (y-5)^2 }{ 16 } = 1\]

Answer 13

= 1?

Answer 14

Yes it's a hyperbola equation

Answer 15

Thought it was -1

Answer 16

Mk lemme think for a moment

Answer 17

x2/a2 − y2/b2 = 1

That's the equation for a hyperbola I think... right?

Answer 18

Think so

Answer 19

K I'm reading a refresher so I can remember how to do this

Answer 20

Okay(:

Answer 21

Okay, so here's what I understand so far:
The hyperbola is centered on (h,k)
So our center point is (4,-5)

Still reading...

Answer 22

Okay!

Answer 23

We need to use:
\[y = \pm \frac{ b }{ a }(x-h) + k\]

Answer 24

Wait I think I got the answer!!

Answer 25

Good, because I just realized I haven't learned this yet and I'm struggling.

Answer 26

I do have the answer though, so I can tell you if it's true or not.

Answer 27

It would be the Vertices: (-1,5),(-7.5); Foci: (-9,5),(1,5)

Answer 28

Yup. Can you explain to me how you got it or do you not have time

Answer 29

I'll just ask a new question so someone can explain it. :D

Answer 30

Well it's a hyperbola that opens left/right so the center is at (h,k) and to find the vertices you use: (h+-a,k), for foci you use (h+-c,k). I was confused on reading it but you helped me!! Oh and the standard equation for that type of hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1