Question

Find the vertices and foci of the hyperbola with equation

(give me a second to write it)

Answer 1

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

Answer 2

Hyperbola Standard Equation:
\[\frac{ (x-h)^{2} }{ a ^{2} } - \frac{ (y-k)^{2} }{ b ^{2} } = 1\]
where (h,k) is the vertex coordinates

Answer 3

That I know

Answer 4

Then you'll have the vertex coordinate
Foci distance = 2a

Answer 5

Vertex is (-4,5)

So the foci are 18 away?

Answer 6

Wait no they are 6 away

Answer 7

Because a is squared

Answer 8

yeah it's 6..

Answer 9

Okay so now what would I do?

Answer 10

@amistre64

Answer 11

well, do we agree that when the x parts zero out, the equation makes no sense?

0 - y^2 = 1

Answer 12

Yes

Answer 13

so the verts are when y=5, since 5-5=0

the x parts are then solved for:

(x+4)^2 = 9

Answer 14

the foci of a hyperB are along the same line as the verts .... so they end in ,5

they also rely on the pythag thrm of the denominators

Answer 15

Okay

Answer 16

tell me what youhave so far

Answer 17

That the vertices will end in 5

Answer 18

verts and focuses, yes

the x part for the verts must satisfy: (x+4)^2/9=1

so solve for x+4 = 3, and x+4 = -3

Answer 19

the focuses we rely on the center, and a distance gotten from pythaging the denominators

Answer 20

Oh so the vertices are

(-1,5)(-7,5)

Answer 21

yes

Answer 22

lets say |f|, the distance a focus is from the center, is equal to:

|f| = sqrt(9+16)

our center is -4 so our x parts for the focuses is: -4 +- sqrt(25)

Answer 23

Got it

Thanks so much, I really appreciate it

Answer 24

youre welcome, and good luck

Answer 25

Thank you :)