Problem 15 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the vertex, focus, and directrix of the parabola. (b) Sketch a graph showing the parabola and its directrix. $$(y+5)^{2}=-6 x+12$$

Step-by-Step Solution

Answer

Vertex: (2, -5); Focus: (0.5, -5); Directrix: x = 3.5.

Step 1: Simplify the Equation

The given equation is \((y+5)^2 = -6x + 12\). First, we simplify the equation by rearranging it to make it look like the standard form of a parabola equation. From \(-6x + 12\), factor out the common term to get \(-6(x - 2)\). Thus the equation becomes \((y+5)^2 = -6(x-2)\).

Step 2: Identify the Form and Key Components of Parabola

The equation \((y - k)^2 = 4p(x - h)\) represents a parabola that opens horizontally. Comparing \((y+5)^2 = -6(x-2)\) to \((y - k)^2 = 4p(x - h)\), we identify the vertex \((h, k)\) as \((2, -5)\) and \(4p = -6\). From \(4p = -6\), solve for \(p\) to get \(p = -\frac{3}{2}\).

Step 3: Find the Focus and Directrix

Since the parabola opens horizontally to the left, find the focus by adding \(p\) to the \(x\)-coordinate of the vertex. Therefore, the focus is located at \((2 + p, -5) = (\frac{1}{2}, -5)\). The directrix is a vertical line given by \(x = h - p\), which is \(x = 2 + \frac{3}{2} = \frac{7}{2}\).

Step 4: Sketch the Parabola, Focus, and Directrix

Plot the vertex \((2, -5)\) on a coordinate plane. Draw the directrix as a vertical line at \(x = \frac{7}{2}\). Plot the focus at \((\frac{1}{2}, -5)\). Sketch the parabola with the vertex and the directrix such that it opens horizontally to the left, indicating the focus as inside and on the axis of the parabola.

Key Concepts

Vertex of a Parabola

When dealing with parabolas, one of the first crucial points to identify is the vertex. The vertex is often thought of as the "tip" or "turning point" of the parabola. In the equation \((y+5)^2 = -6(x-2)\), the vertex can be easily determined by comparing it to the standard form of a parabola that opens horizontally, \((y-k)^2 = 4p(x-h)\). Here, \(h\) and \(k\) are the coordinates of the vertex. From this equation, we identify the vertex as \( (2, -5) \). The vertex helps us understand the symmetry of the parabola and provides a reference point for plotting.

• Vertex: The point \( (2, -5) \) is where the parabola changes direction.
• Role: Helps in understanding the orientation of the parabola.

A vertex is crucial not only for graphing but also in determining the equation when points are known.

Focus of a Parabola

The focus of a parabola is a point that lies inside the parabola itself. It is unique because any point on the parabola is equidistant from the focus and the directrix. For our parabola given by \((y+5)^2 = -6(x-2)\), the parameter \(p\) is calculated from the equation factor \(4p = -6\), so \(p = -\frac{3}{2}\). To find the focus, we add \(p\) to the \(x\)-coordinate of the vertex. This step results in a focus located at \( (\frac{1}{2}, -5) \). Understanding the position of the focus helps in the accurate sketching of the parabola.

• Focus: Placed at \( (\frac{1}{2}, -5) \).
• Purpose: Determines the degree of "openness" of the parabola and aids in proper graphing.

Knowing the focus allows us to better understand how the parabola behaves and its geometric properties.

Directrix of a Parabola

Another essential feature of a parabola is its directrix. The directrix is a line, rather than a point, which plays a role in defining the parabola. It is perpendicular to the axis of the parabola and lies outside the parabola. From the equation \((y+5)^2 = -6(x-2)\), we've calculated \(p\) as \(-\frac{3}{2}\), and thus for a parabola opening horizontally to the left, the directrix becomes evident. The calculated directrix is a vertical line at \(x = 2 + \frac{3}{2} = \frac{7}{2}\). This means it lies at \(x = \frac{7}{2}\), balancing the position of the focus.

• Directrix: A vertical line expressed as \(x = \frac{7}{2}\).
• Significance: Each point on the parabola is equally distant from the focus and this line.

The directrix serves as a tool for defining the parabola's "openness" and maintains alignment with the focus, ensuring the curve is correctly shaped.