Problem 22 - Algebra and Trigonometry

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, 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 at (-3, 0), focus (-3, 3), and directrix y = -3.

Step 1: Rearrange Equation

First, we rearrange the terms of the given equation, grouping the terms involving the same variable. The equation is \(x^2 + 6x + 12y + 9 = 0\). We can rewrite it as \(x^2 + 6x = -12y - 9\).

Step 2: Complete the Square for x

To complete the square for the \(x\) terms, take the linear coefficient (6), divide it by 2, and square it. This gives \((\frac{6}{2})^2 = 9\). Add and subtract 9 to the left side: \(x^2 + 6x + 9 - 9 = -12y - 9\). This can be simplified to \((x+3)^2 - 9 = -12y - 9\).

Step 3: Simplify Equation

Add 9 to both sides to move the constants: \((x+3)^2 = -12y\). Divide both sides by -12 to isolate \(y\): \(y = -\frac{1}{12}(x+3)^2\).

Step 4: Identify the Conic Section

The equation \(y = -\frac{1}{12}(x+3)^2\) is in the form of a parabola with the equation \(y - k = a(x-h)^2\). Here, the parabola opens downward because the coefficient in front of the \((x+3)^2\) term is negative.

Step 5: Find the Vertex of the Parabola

The vertex of the parabola can be identified from the equation \(y = -\frac{1}{12}(x+3)^2\) as \((-3, 0)\), because it is in the form \(y = a(x-h)^2 + k\), where \(h = -3\) and \(k = 0\).

Step 6: Find the Focus of the Parabola

For the parabola \(y = a(x-h)^2 + k\), the focus is located at \((h, k + \frac{1}{4a})\). Substituting \(a = -\frac{1}{12}\), \(h = -3\), and \(k = 0\), the focus is \((-3, 0 + \frac{1}{4(-1/12)})\), which simplifies to \((-3, 3)\).

Step 7: Find the Directrix of the Parabola

The equation for the directrix is \(y = k - \frac{1}{4a}\). Substituting the values \(a = -\frac{1}{12}\) and \(k = 0\), we find \(y = 0 - \frac{1}{4(-1/12)} = -3\). Thus, the directrix is \(y = -3\).

Step 8: Sketch the Parabola

The parabola opens downward with vertex at \((-3, 0)\), focus \((-3, 3)\), and directrix \(y = -3\). Draw the axis of symmetry \(x = -3\), and sketch the parabola so it opens downward, passing through the vertex and moving away from the directrix.

Key Concepts

Complete the Square

Completing the square is a fundamental algebraic method used to transform a quadratic equation into a more useful form. The process involves creating a perfect square trinomial from a quadratic expression.

When tackling a quadratic in the form of \(ax^2 + bx + c = 0\), completing the square allows us to rewrite this equation so that it contains a binomial squared. This is extremely helpful in studying conic sections. For example, if you start with an equation like \((x^2 + 6x)\), you can turn it into a perfect square by determining what number completes the square: divide the coefficient of \(x\) by 2 and then square it. In this case, \((\frac{6}{2})^2 = 9\).

Add and subtract this value within the equation to keep it balanced, and you'll have \((x+3)^2\) followed by any necessary correction of terms.

Conic Sections

Conic sections, as the name suggests, are curves obtained by intersecting a plane with a cone. Depending on the angle and position of this intersection, we classify conic sections into circles, ellipses, parabolas, and hyperbolas. Each has its own set of unique properties.

In our exercise, we encountered a parabola. Parabolas are one of the simplest conic sections and are recognized by their U-shape. The equation of a parabola, typically given as \(y = ax^2 + bx + c\), can be manipulated by completing the square to yield an equation that highlights its vertex form. This is represented as \(y = a(x-h)^2 + k\) where \( (h, k)\) is the vertex.

By understanding the geometric significance of parabolas among conic sections, we can identify key characteristics like the focus and directrix, which play a vital role in defining the parabola's shape and position.

Vertex of a Parabola

The vertex of a parabola is the point where it changes direction, marking either its highest or lowest point. This is critical in graphing or analyzing the parabola's behavior.

In the standard vertex form \(y = a(x-h)^2 + k\), the vertex is simply the point \( (h, k)\). From our specific equation \((y = -\frac{1}{12}(x+3)^2)\), the vertex is at \((-3, 0)\). This information is pivotal as it provides the parabola's axis of symmetry, dividing it into two mirror-image halves.

Understanding the vertex helps in determining where other critical elements like the focus and directrix will be located, providing a comprehensive view of the parabola's structure and orientation.

Focus of a Parabola

The focus of a parabola is a fixed point located inside the curve such that any point on the parabola is equidistant from the focus and the directrix, a linear guide below or above the curve.

For a parabola given by \(y = a(x-h)^2 + k\), the focus can be found at \((h, k+\frac{1}{4a})\). In our problem, substituting the values \(a = -\frac{1}{12}\), \(h = -3\), and \(k = 0\), leads us to the focus at point \((-3, 3)\).

The position of the focus is essential for visualizing how the parabola curves. The vertex always falls exactly halfway between the focus and the directrix, with the parabola opening around it.

Directrix of a Parabola

The directrix of a parabola is a straight line that, together with the focus, helps to define and shape the curve of the parabola. It is always positioned perpendicularly to the axis of symmetry, and the parabola is equidistant from any point along its edge to both the directrix and the focus.

This line can be described by the formula \(y = k - \frac{1}{4a}\), derived from the vertex form of a parabola's equation. For our example, with given parameters \(a = -\frac{1}{12}\) and \(k = 0\), the directrix is found at \(y = -3\).

Its placement is crucial since it determines the orientation and widening of the parabola. As the curve mirrors the directrix from the focus, understanding the directrix helps clarify the geometrics in the sketching and conceptualization of parabolic shapes.