- anonymous

Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity of y minus one squared divided by one hundred and forty four = 1

- chestercat

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- anonymous

- anonymous

waddup?

- anonymous

\[\frac{(x-5)^2}{81}-\frac{(y-1)^2}{4}=1\]?

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## More answers

- anonymous

this is way easier than you think
but you need to know two things first
a) the center and
b) which way it is oriented

- anonymous

I know that the hyperbolas are oriented left/right

- anonymous

i thought ya all that earlier

- anonymous

taught*

- anonymous

center?

- anonymous

@satellite73 you mean what i taught him?

- anonymous

(0,5) ?

- anonymous

My bad, (5, 1)

- anonymous

yes

- anonymous

Ok so now what can I do?

- anonymous

find c

- anonymous

\[\frac{(x-5)^2}{81}-\frac{(y-1)^2}{4}=1\]\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]
what is \(a\)?

- anonymous

Ok. c = 15

- anonymous

a = 81, b =144

- anonymous

no

- anonymous

no

- anonymous

\[a^2=81,b^2=144\]

- anonymous

\[c^2=a^2+b^2\]

- anonymous

thats why i told ya to put em into ^2 form

- anonymous

Oh sorry. a = 9, b = 12

- anonymous

Ok, so c = 15 because c^2 = 225

- anonymous

- anonymous

yea?

- anonymous

you know the hyperbola is transverse in x-axis so ( center +/- c, center) is your foci

- anonymous

and same for the vertices

- anonymous

Wait. These are my choices:
A. Vertices: (17, 1), (-7, 1); Foci: (-7, 1), (17, 1)
B. Vertices: (14, 1), (-4, 1); Foci: (-10, 1), (20, 1)
C. Vertices: (1, 14), (1, -4); Foci: (1, -10), (1, 20)
D. Vertices: (1, 17), (1, -7); Foci: (1, -7), (1, 17)
So I know it has to be between A and B.

- anonymous

Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7).
If my answer choices are:
A. y squared over nine minus x squared over forty = 1
B. y squared over forty minus x squared over nine = 1
C. y squared over forty nine minus x squared over nine = 1
D. y squared over nine minus x squared over forty nine = 1
I was thinking it was between B and D.

- anonymous

lemme check

- anonymous

you see foci is c^2 = a^2+b^2
so we get 7^2 - 3^2 = 40

- anonymous

Oh ok. So that would make it A.

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