Step-by-step Solution:

Given Equation: \[ 3x^2 + 10xy + 11y^2 + 12y + 5 = 0 \] The equation can be compared to the general second-degree equation: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \] From the given equation: - \(a = 3\) (coefficient of \(x^2\)) - \(2h = 10 \implies h = 5\) (coefficient of \(xy\)) - \(b = 11\) (coefficient of \(y^2\)) - \(g = 0\) (as there is no \(x\)-term) - \(2f = 12 \implies f = 6\) (coefficient of \(y\)) - \(c = 5\) (constant term) The determinant of the conic section helps identify its type. It is given by: \[ h^2 - ab \] Substitute the values: \[ h^2 - ab = 5^2 - (3)(11) \] \[ h^2 - ab = 25 - 33 = -8 \] - If \(h^2 - ab < 0\), the conic is an ellipse. - Here, since \(h^2 - ab = -8 < 0\), the given equation represents an ellipse.