anonymous
  • anonymous
how do you create an equation of an ellipse with a major axis of length 18 and foci located at (4,7) and (4,11)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
do you know what this looks like? we need that first
anonymous
  • anonymous
an ellipse, but how can you tell if its horizontal or vertical?
anonymous
  • anonymous
because of the foci the \(x\) values are the same, whereas the \(y\) values are different

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anonymous
  • anonymous
|dw:1378431271146:dw|
anonymous
  • anonymous
|dw:1378431302034:dw|
anonymous
  • anonymous
so it is going to look like \[\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1\] you need the center, which is half way between \((4,7)\) and \((4,11)\) and also \(a\) and \(b\)
anonymous
  • anonymous
you got the center?
anonymous
  • anonymous
(4,9)
anonymous
  • anonymous
k good so now we are here \[\frac{(y-9)^2}{a^2}+\frac{(x-4)^2}{b^2}=1\]
anonymous
  • anonymous
how far from the center to the top part of the ellipse?
anonymous
  • anonymous
9
anonymous
  • anonymous
k good, so now we know \(a=9\) and so \(a^2=81\)
anonymous
  • anonymous
now we are here \[\frac{(y-9)^2}{9^2}+\frac{(x-4)^2}{b^2}=1\]
anonymous
  • anonymous
how far from the center to the foci?
anonymous
  • anonymous
2
anonymous
  • anonymous
b^2=4?
anonymous
  • anonymous
got it so \(c=2\) and therefore \(c^2=4\) you need \(b^2=a^2-c^2\)
anonymous
  • anonymous
no that was \(c\) not \(b\) we still have to find \(b\) or actually \(b^2\)
anonymous
  • anonymous
b^2 = the square root of 77?
anonymous
  • anonymous
i think actually \(b^2=77\) not \(b=77\)
anonymous
  • anonymous
I get it! thank you for all the help now I understand it.
anonymous
  • anonymous
lets check looks good http://www.wolframalpha.com/input/?i=ellipse+%28y-9%29^2%2F81%2B%28x-4%29^2%2F77%3D1\yw

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