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watson3
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find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0
 one year ago
 one year ago
watson3 Group Title
find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0
 one year ago
 one year ago

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Loser66 Group TitleBest ResponseYou've already chosen the best response.0
ok, sorry, backward, perfect square first, then isolate the number
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
hey friend, I help only, I don't do it for you.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
\[\large 4x^2+9y^218y27=0\] Complete the square with the yterms. In order to do that, we must turn the "9y^2" into y^2. DIVIDE both sides by 9. \[\large \frac{4}{9}x^2+y^22y3=0\] 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: \[\huge y^22y=3\] Complete the square. \[\large (y1)^2=3+1\] \[\large (y1)^2=4\] Now bring back the (4/9)x^2. SO now we have the equation: \[\frac{4}{9}x^2+(y1)^2=4\] Now we have to divide both sides by 4. We will have: \[\huge \frac{x^2}{9}+\frac{(y1)^2}{4}=1\] 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: \[\huge b^2=a^2(1e^2)\] \[\large 4=99e^2\] \[\large e^2=\frac{5}{9}\] \[\large e=\frac{\sqrt{5}}{3}\] Therefore the coordinates of the foci is: \[\huge (\sqrt{5}, 0)\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
\[\huge (\pm \sqrt{5}, 0)\] Sorry about that. This website is a bit too overwhelming for my computer to handle. Forgot the plus or minus sign.
 one year ago
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