anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@saseal
anonymous
  • anonymous
waddup?
anonymous
  • anonymous
\[\frac{(x-5)^2}{81}-\frac{(y-1)^2}{4}=1\]?

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More answers

anonymous
  • 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
  • anonymous
I know that the hyperbolas are oriented left/right
anonymous
  • anonymous
i thought ya all that earlier
anonymous
  • anonymous
taught*
anonymous
  • anonymous
center?
anonymous
  • anonymous
@satellite73 you mean what i taught him?
anonymous
  • anonymous
(0,5) ?
anonymous
  • anonymous
My bad, (5, 1)
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok so now what can I do?
anonymous
  • anonymous
find c
anonymous
  • 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
  • anonymous
Ok. c = 15
anonymous
  • anonymous
a = 81, b =144
anonymous
  • anonymous
no
anonymous
  • anonymous
no
anonymous
  • anonymous
\[a^2=81,b^2=144\]
anonymous
  • anonymous
\[c^2=a^2+b^2\]
anonymous
  • anonymous
thats why i told ya to put em into ^2 form
anonymous
  • anonymous
Oh sorry. a = 9, b = 12
anonymous
  • anonymous
Ok, so c = 15 because c^2 = 225
anonymous
  • anonymous
@saseal
anonymous
  • anonymous
yea?
anonymous
  • anonymous
you know the hyperbola is transverse in x-axis so ( center +/- c, center) is your foci
anonymous
  • anonymous
and same for the vertices
anonymous
  • 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
  • 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
  • anonymous
lemme check
anonymous
  • anonymous
you see foci is c^2 = a^2+b^2 so we get 7^2 - 3^2 = 40
anonymous
  • anonymous
Oh ok. So that would make it A.

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