sloppycanada
  • sloppycanada
Ellipses http://gyazo.com/e897fc96e592ab1e5dcba825538b6d7e y^2/100 - x^2/16 = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
no not this time
anonymous
  • anonymous
center is not \((0,0)\) find the center first
sloppycanada
  • sloppycanada
(1,2)?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
so it is going to look like \[\frac{(x-1)^2}{a^2}-\frac{(y-2)^2}{b^2}=1\]
anonymous
  • anonymous
you need \(a\) and \(b\)
sloppycanada
  • sloppycanada
But wouldn't x and y be switched? Since the graph is going a different way?
anonymous
  • anonymous
no
sloppycanada
  • sloppycanada
So it's not like an ellipses that way.
anonymous
  • anonymous
|dw:1435717476572:dw|
anonymous
  • anonymous
scuse the crappy pictures
sloppycanada
  • sloppycanada
No worries, mine are worse Okay so for this graph it'd be - (x^2-2)/100 - (y^2-1)/16 = 1
sloppycanada
  • sloppycanada
Also for this one - you said it was x than y. http://gyazo.com/99da097eae7203d795aaab65586d4f48
anonymous
  • anonymous
then i made a mistake
anonymous
  • anonymous
also no, not 100 and 16
anonymous
  • anonymous
the distance from the center to the vertex is 5
anonymous
  • anonymous
not the whole length from one vertex to the other, which is 10 so \(25\) goes under the \(x\) part
anonymous
  • anonymous
similarly under the \(y\) part put a 4, not 16
sloppycanada
  • sloppycanada
(x^2-2)/25 - (y^2-1)/4 = 1 is my final answer?
anonymous
  • anonymous
yes want to check it?
anonymous
  • anonymous
oh no !!
sloppycanada
  • sloppycanada
No?
anonymous
  • anonymous
not \(x^2-1\) up top but rather \((x-1)^2\)
anonymous
  • anonymous
sorry \((x-2)^2\)
sloppycanada
  • sloppycanada
(x-2)^2/25 - (y-1)^2/4 = 1
anonymous
  • anonymous
here is the check that \[\frac{(x-2)^2}{25}-\frac{(y-1)^2}{4}=1\] is right http://www.wolframalpha.com/input/?i=hyperbola+%28x-1%29^2%2F25-%28y-2%29^2%2F4%3D1
anonymous
  • anonymous
hard to tell from the graphs, but you can see from the vertices that it is correct
sloppycanada
  • sloppycanada
Okay thanks!
sloppycanada
  • sloppycanada
So many hyperbola questions...
anonymous
  • anonymous
there is a certain sameness to them

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