To identify the conic section represented by the equation \(3x^2 + 2y^2 - 24x + 30 = 0\), we can follow the steps to rearrange and analyze it.
First, we can rearrange the equation into a more standard form. Let's group the terms related to \(x\) and \(y\):
\[ 3x^2 - 24x + 2y^2 + 30 = 0 \]
Next, we can isolate the terms involving \(x\) and \(y\):
\[ 3x^2 - 24x + 2y^2 = -30 \]
Now, we will complete the square for the \(x\) terms. The \(x\) terms are \(3x^2 - 24x\). To complete the square, factor out the coefficient of \(x^2\) (which is 3):
\[ 3(x^2 - 8x) \]
Now, complete the square inside the parentheses. Take half of \(-8\) (which is \(-4\)), square it (getting 16), and then add and subtract it inside the parentheses:
\[ 3(x^2 - 8x + 16 - 16) = 3((x - 4)^2 - 16) = 3(x - 4)^2 - 48 \]
Substituting this back into the equation gives us:
\[ 3(x - 4)^2 - 48 + 2y^2 = -30 \]
Now, move \(-48\) to the right-hand side:
\[ 3(x - 4)^2 + 2y^2 = -30 + 48 \]
This simplifies to:
\[ 3(x - 4)^2 + 2y^2 = 18 \]
Now divide the entire equation by 18 to put it in standard form:
\[ \frac{3(x - 4)^2}{18} + \frac{2y^2}{18} = 1 \]
Simplifying this gives:
\[ \frac{(x - 4)^2}{6} + \frac{y^2}{9} = 1 \]
This equation is now in the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \((h, k)\) is the center of the ellipse, and \(a^2\) and \(b^2\) are the squares of the semi-axis lengths.
Since both terms are positive and the equation resembles that of an ellipse, we conclude that the conic section represented by the original equation \(3x^2 + 2y^2 - 24x + 30 = 0\) is an ellipse.