anonymous
  • anonymous
Will Medal, just really need help!! Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 3.x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
looks like you are stuck with analytic geometry :) now lets start with how standard form looks like\[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]
anonymous
  • anonymous
Ok @saseal
anonymous
  • anonymous
|dw:1439304877703:dw|

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anonymous
  • anonymous
My answer choices are: A. y squared over 144 minus x squared over 16 = 1 B. y squared over 16 minus x squared over 36 = 1 C. y squared over 16 minus x squared over 144 = 1 D. y squared over 36 minus x squared over 4 = 1 I'm between B and C
anonymous
  • anonymous
now we have the gradient y=a/b which is 1/3
anonymous
  • anonymous
and we know a is 4 so b gotta be 12 since the gradient have to match up
anonymous
  • anonymous
you can guess which one it is by now :)
anonymous
  • anonymous
C? @saseal
anonymous
  • anonymous
yes
anonymous
  • anonymous
Can you help me with a few others?
anonymous
  • anonymous
ok
anonymous
  • anonymous
Thanks! @saseal Find the center, vertices, and foci of the ellipse with equation x squared divided by 400 plus y squared divided by 625 = 1.
anonymous
  • anonymous
\[\frac{ x^2 }{ 400 } + \frac{ y^2 }{ 625 } = 1\] this should be easy by now
anonymous
  • anonymous
\[\frac{ x^2 }{ 20^2 } + \frac{ y^2 }{ 25^2 } = 1\]
anonymous
  • anonymous
where do you think is the major axis of this egg?
anonymous
  • anonymous
major axis is the longer side
anonymous
  • anonymous
The y-axis
anonymous
  • anonymous
correct
anonymous
  • anonymous
now think where's the center
anonymous
  • anonymous
theres no h and k here
anonymous
  • anonymous
0,0
anonymous
  • anonymous
there we have the both the center and vertices now
anonymous
  • anonymous
What's the vertices though?
anonymous
  • anonymous
center (0, 0) vertices (0, ±25)
anonymous
  • anonymous
Oh ok!
anonymous
  • anonymous
|dw:1439306628656:dw|
anonymous
  • anonymous
now we need to find c to get the foci
anonymous
  • anonymous
\[c=\sqrt{a^2-b^2}\]
anonymous
  • anonymous
Ok, so C = 15? @saseal
anonymous
  • anonymous
+/- 15
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok so would my answer be: Center: (0, 0); Vertices: (-25, 0), (25, 0); Foci: (-15, 0), (15, 0)
anonymous
  • anonymous
Or no. It would be: Center: (0, 0); Vertices: (0, -25), (0, 25); Foci: (0, -15), (0, 15
anonymous
  • anonymous
Could you help me with one more? @saseal
anonymous
  • anonymous
yea
anonymous
  • anonymous
sry im not looking at the pc all the time
anonymous
  • anonymous
No problem. And I'm in a new chapter so: Find the derivative of f(x) = 6 divided by x at x = -2.
anonymous
  • anonymous
And also: Find the derivative of f(x) = 4x + 7 at x = 5. @saseal
anonymous
  • anonymous
sry i fell asleep

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