91Ó°ÊÓ

We will rewrite the given equation \((x+3)^2 = -2(y-4)\) into the standard form for conic sections, which is: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). In this case, our equation is: \((x+3)^2 = -2(y-4)\) Expanding the equation gives us: \(x^2 + 6x + 9 = -2(y - 4)\) Next, expand the -2: \(x^2 + 6x + 9 = -2y + 8\) Now, move all the terms to the left side to get the equation in standard form: \(x^2 + 6x + 2y + 1 = 0\) Hence, our equation is now \(x^2 + 6x + 2y + 1 = 0\).

Now, let's identify the type of conic section the equation represents. A conic section can be a circle, ellipse, parabola, or hyperbola, depending on the values of A, B, and C. We know their general forms as follows: 1. Circle or Ellipse: \(Ax^2 + Cy^2 + Dx + Ey + F = 0 \) where \(A = C\) and \(B = 0\) 2. Parabola: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) where \(B^2 = 4AC\) 3. Hyperbola: \(Ax^2 - Cy^2 + Dx + Ey + F = 0\) where \(A = -C\) and \(B = 0\) In our equation, we have: \(A = 1\), \(B = 0\), and \(C = 2\) As we have \(B = 0\) and \(A \neq C\), it means our conic section is neither a circle nor a hyperbola. It also can't be a parabola as \(B^2 \neq 4AC\). Therefore, our conic section is an ellipse. Hence, based on our identification, the given equation \((x+3)^2 = -2(y-4)\) represents an ellipse. The student should match this equation with the ellipse conic section option among the choices (a)-(d).