## watson3 2 years ago find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0

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1. Loser66

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

2. Loser66

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

3. Azteck

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

4. Azteck

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