anonymous
  • anonymous
find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0
Precalculus
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
ok, sorry, backward, perfect square first, then isolate the number
Loser66
  • Loser66
hey friend, I help only, I don't do it for you.
anonymous
  • anonymous
\[\large 4x^2+9y^2-18y-27=0\] Complete the square with the y-terms. In order to do that, we must turn the "9y^2" into y^2. DIVIDE both sides by 9. \[\large \frac{4}{9}x^2+y^2-2y-3=0\] Move the 3 to the otherside and let's get rid of the (4/9)x^2 for a moment. We will add that back into the equation afterwards. So we have this: \[\huge y^2-2y=3\] Complete the square. \[\large (y-1)^2=3+1\] \[\large (y-1)^2=4\] Now bring back the (4/9)x^2. SO now we have the equation: \[\frac{4}{9}x^2+(y-1)^2=4\] Now we have to divide both sides by 4. We will have: \[\huge \frac{x^2}{9}+\frac{(y-1)^2}{4}=1\] The equation of the foci when a>b is: \[\huge (\pm ae, 0)\] In this ellipse, the value of a=3 and b=2 To find "e", we use the equation: \[\huge b^2=a^2(1-e^2)\] \[\large 4=9-9e^2\] \[\large e^2=\frac{5}{9}\] \[\large e=\frac{\sqrt{5}}{3}\] Therefore the coordinates of the foci is: \[\huge (\sqrt{5}, 0)\]

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anonymous
  • anonymous
\[\huge (\pm \sqrt{5}, 0)\] Sorry about that. This website is a bit too overwhelming for my computer to handle. Forgot the plus or minus sign.

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