Problem 50 - Precalculus: Mathematics for Calculus

Question

Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an cllipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$x^{2}+6 x+12 y+9=0$$

Step-by-Step Solution

Answer

The equation represents a downward-opening parabola with vertex (-3, 0), focus (-3, -3), and directrix y = 3.

Step 1: Group and Rearrange Terms

First, we need to rearrange the equation to make completing the square easier. Start by moving the terms around so that all terms in one variable are grouped together. Rewrite the equation as follows: \[x^2 + 6x = -12y - 9\].

Step 2: Complete the Square in the x-term

To complete the square for the \(x\)-terms, we take the coefficient of \(x\), which is 6, divide it by 2, and then square it. This gives us \((\frac{6}{2})^2 = 9\). Add and subtract this 9 in the equation: \[x^2 + 6x + 9 - 9 = -12y - 9\].

Step 3: Simplify the Equation

Now, factor the completed square trinomial and simplify the equation: \[(x+3)^2 - 9 = -12y - 9\]. Add 9 to both sides to balance the equation: \[(x+3)^2 = -12y\].

Step 4: Rewrite in Standard Form

Divide the entire equation by -12 to isolate \(y\): \[y = -\frac{1}{12}(x+3)^2\]. This resembles the equation of a parabola \(y = ax^2\), indicating that it's a parabola opening downward.

Step 5: Determine Key Features of the Parabola

Since \(y = -\frac{1}{12}(x+3)^2\), the vertex is at \((-3, 0)\). The focus can be found using \(a = -\frac{1}{12}\), where \(a = \frac{1}{4f}\), so \(f = -3\). Thus, the focus is at \((-3, -3)\). The directrix is found at \(y = f + p\), so it is \(y = 3\).

Step 6: Sketch the Graph

Plot the vertex at \((-3, 0)\), the focus at \((-3, -3)\), and draw the directrix line at \(y=3\). Sketch a parabola opening downward that mirrors over the \(x = -3\) axis, with the vertex and focus guiding the curvature.

Key Concepts

Parabola

A parabola is a specific type of conic section that is shaped like a symmetrical open curve. These curves are defined as the set of all points that are equidistant from a fixed point, known as the focus, and a fixed straight line, called the directrix. Parabolas can open in various directions: upwards, downwards, left, or right, depending on the equation's form.

In our given equation, once it's rearranged and simplified to the parabola standard form \(y = -\frac{1}{12}(x+3)^2\), it indicates a downward-opening parabola. The equation helps determine how wide or narrow the parabola is and in which direction it opens. This understanding is crucial in solving conic section problems since it leads us directly to the important features like the focus and the vertex.

Completing the Square

Completing the square is a method used in algebra to transform a quadratic equation into a perfect square trinomial. This process involves adding and subtracting a particular number to make the square of a binomial.

Here's how it applies to our example: Start with the expression \(x^2 + 6x\). To complete the square, first take the coefficient of \(x\), which is 6, divide by 2, and square the result to get 9. We then rewrite the expression as \( (x+3)^2 - 9 \). This makes it easier to handle because it's now a perfect square trinomial (a binomial squared), which is simpler to analyze and graph.

This technique is especially useful when converting conic sections from their general form to a more recognizable standard form. It allows us to easily identify the critical features of the conic, such as the vertex of a parabola, by visually simplifying the equation.

Vertex Focus Directrix

Understanding the vertex, focus, and directrix of a parabola is essential in graphing and analyzing its properties:

• Vertex: The vertex is the peak or the lowest point of the parabola, depending on its direction. For our parabola \(y = -\frac{1}{12}(x+3)^2\), the vertex is at \((-3, 0)\). This is found from the transformation completed in the square method, simplifying the complex equation into a form that reveals the vertex directly as \( (x+3)^2\).


• Focus: The focus is a point inside the parabola where all the lines reflect off the curve and converge. It's computed using the relation \(a = \frac{1}{4f}\). Since \(a = -\frac{1}{12}\), solve for \(f\) and find the focus at \((-3, -3)\).


• Directrix: The directrix is the line opposite the parabola's opening that helps in reflecting lines to the focus. It's given as \(y = 3\), creating a clear point of symmetry along the curve.

These components show how a parabola is geometrically structured and enable precise sketching and understanding of its path and boundary.