Problem 37 - Precalculus with Limits

Question

In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \(x^2 + 6y = 0\)

Step-by-Step Solution

Answer

The vertex of the parabola is at (0,0), the focus is at (0,-1/24), and the equation of the directrix is \(y = 1/24\). The parabola opens downwards.

Step 1: Convert into standard form

First rewrite the equation into the standard form: Divide all terms by 6 to isolate y on one side, resulting in \(x^2/6 = -y\).

Step 2: Find the value of p

Compare the obtained equation with the standard form \(x^2 = 4py\). Here, \(4p = 1/6\). Solving this for p, we find that p = -1/24. Note: The negative sign indicates that our parabola will open downwards.

Step 3: Find the vertex

The vertex of a parabola in the form \(x^2 = 4py\) is always at the origin (0,0) since the equation is not shifted.

Step 4: Find the focus

The formula for the focus of the parabola in the form \(x^2 = 4py\) is (0,p). Substituting for p, the focus would be at (0,-1/24).

Step 5: Find the directrix

The formula for the directrix of a parabola in the form \(x^2 = 4py\) is \(y = -p\). Substituting for p, the directrix is \(y = 1/24\).

Step 6: Sketch the graph

The parabola is downward-opening with the vertex at the origin. The focus is inside the parabola below the vertex and the directrix is a horizontal line above the vertex.

Key Concepts

Standard Form of a Parabola

Understanding the standard form of a parabola is crucial for graphing and analyzing its properties. A parabola's standard form is represented by the equation \(y = ax^2 + bx + c\) for a vertical parabola, or \(x = ay^2 + by + c\) for a horizontal parabola, where \(a\), \(b\), and \(c\) are coefficients that determine the shape and position of the parabola.

In the context of the given exercise, the equation \(x^2 + 6y = 0\) was transformed into a standard form by dividing all terms by 6, resulting in \(x^2/6 = -y\). This indicates that the parabola opens downwards because the y-term is negative. By further comparing this with the general standard form \(x^2 = 4py\), we can find the value of \(p\) which plays a fundamental role in determining the characteristics of the parabola such as the focus and directrix.

Vertex of a Parabola

The vertex of a parabola is a critical point where the curve changes direction, and it's also the minimum or maximum point depending on whether the parabola opens upwards or downwards. For parabolas in standard form \(x^2 = 4py\) or \(y^2 = 4px\), the vertex is at the origin (0,0) when the equation lacks a horizontal or vertical shift.

With the exercise's example of \(x^2 + 6y = 0\), or \(x^2/6 = -y\), the parabola is not shifted from the origin, so the vertex remains at (0,0). Identifying the vertex is a pivotal step in graphing the parabola as it provides a reference point from which the rest of the parabola's shape can be determined.

Focus of a Parabola

The focus of a parabola is a point located inside the curve from which all points on the parabola are equidistant from a given line called the directrix. The focus plays a vital role in the reflective properties of a parabola. For a parabola defined by \(x^2 = 4py\), the focus is given by the coordinate (0,p).

In our exercise, we determined the value of \(p\) to be -1/24, which suggests that the focus lies below the vertex at the point (0,-1/24) because the parabola opens downwards. The focus of a parabola is important not only in graphing but also in applications such as satellite dishes and flashlights, where the reflecting property of parabolas is exploited.

Directrix of a Parabola

The directrix of a parabola is a straight line that, together with the focus, defines the symmetric property of a parabola. For each point on the parabola, the distance to the focus is equal to the distance to the directrix. In the equation \(x^2 = 4py\), the directrix is represented by the line \(y = -p\).

Using the value of \(p\) found in the exercise, -1/24, the directrix of the parabola is the line \(y = 1/24\). This line is horizontal and is located above the vertex since the parabola opens downwards. When graphing a parabola, the directrix serves as a reference for ensuring that the distance relationship is maintained, thus preserving the parabolic shape.