Question

find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0

Answer 1

ok, sorry, backward, perfect square first, then isolate the number

Answer 2

hey friend, I help only, I don't do it for you.

Answer 3

\[\large 4x^2+9y^2-18y-27=0\]
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:
\[\huge (\sqrt{5}, 0)\]

Answer 4

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