To rewrite the given equation of the ellipse \(3x^2 + 2y^2 + 18x + 20y + 65 = 0\) in standard form, we first need to rearrange and complete the square for both \(x\) and \(y\). Here are the steps:
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Rearranging the equation: \[ 3x^2 + 18x + 2y^2 + 20y + 65 = 0 \] Move the constant to the other side: \[ 3x^2 + 18x + 2y^2 + 20y = -65 \]
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Completing the square for \(x\):
- Factor out the coefficient of \(x^2\) from the \(x\) terms: \[ 3(x^2 + 6x) + 2y^2 + 20y = -65 \]
- To complete the square for \(x^2 + 6x\):
- Take half of 6, square it: \( (3)^2 = 9 \). \[ 3(x^2 + 6x + 9 - 9) + 2y^2 + 20y = -65 \] \[ 3((x + 3)^2 - 9) + 2y^2 + 20y = -65 \] \[ 3(x + 3)^2 - 27 + 2y^2 + 20y = -65 \] Now, move -27 to the right: \[ 3(x + 3)^2 + 2y^2 + 20y = -38 \]
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Completing the square for \(y\): \[ 2(y^2 + 10y) = -38 \]
- Take half of 10, square it: \( (5)^2 = 25 \). \[ 2(y^2 + 10y + 25 - 25) = -38 \] \[ 2((y + 5)^2 - 25) = -38 \] \[ 2(y + 5)^2 - 50 = -38 \] Now, move -50 to the right: \[ 3(x + 3)^2 + 2(y + 5)^2 = 12 \]
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Dividing by 12 to get standard form: \[ \frac{3(x + 3)^2}{12} + \frac{2(y + 5)^2}{12} = 1 \] Simplify: \[ \frac{(x + 3)^2}{4} + \frac{(y + 5)^2}{6} = 1 \]
Thus, the standard form of the ellipse is: \[ \frac{(x + 3)^2}{4} + \frac{(y + 5)^2}{6} = 1 \]
The correct response from the options provided is: \[ \frac{(x + 3)^2}{4} + \frac{(y + 5)^2}{6} = 1 \]
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