Problem 11 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$x^{2}=8 y$$

Step-by-Step Solution

Answer

Focus: (0, 2), Directrix: y = -2, Focal Diameter: 8.

Step 1: Identify the form of the parabola equation

The given equation is \(x^2 = 8y\). This equation can be rewritten in the form \(x^2 = 4py\), which is the standard form of a vertical parabola. Our goal is to identify the parameter \(p\) from this form, which will be essential in finding the focus, directrix, and focal diameter.

Step 2: Determine the value of 'p'

In the equation \(x^2 = 4py\), the term \(4p\) corresponds with the coefficient of \(y\). By comparing it with our equation \(x^2 = 8y\), we find that \(4p = 8\). Therefore, \(p = 2\).

Step 3: Find the focus of the parabola

For a parabola \(x^2 = 4py\), the vertex is at the origin \((0,0)\). The focus is located at \((0, p)\), given that the parabola opens upwards. Using \(p = 2\), the focus of this parabola is at \((0, 2)\).

Step 4: Determine the directrix of the parabola

The directrix of a parabola in standard form \(x^2 = 4py\) is a horizontal line at \(y = -p\). With \(p = 2\), the directrix of this parabola is the line \(y = -2\).

Step 5: Calculate the focal diameter

The focal diameter is the length of the line segment that passes through the focus and is perpendicular to the axis of symmetry of the parabola. This is equal to the absolute value of \(4p\). Given \(p = 2\), the focal diameter is \(\lvert 4 \times 2 \rvert = 8\).

Step 6: Sketch the graph of the parabola and directrix

Plot the vertex at the origin \((0,0)\), the focus at \((0, 2)\), and the directrix as the horizontal line \(y = -2\). Sketch the parabola shape opening upwards, ensuring it is symmetric with respect to the y-axis and highlights the position of the focus and directrix. The parabola should appear to get wider as it moves away from the vertex.

Key Concepts

Focus of a Parabola

A parabola is a symmetric curve that has a special point called the "focus." This point helps determine the shape and direction of the parabola.

The focus of a parabola is located at \(0, p\) for a vertical parabola in the standard form \(x^2 = 4py\).
To find the focus, you need to identify the value of \(p\).

• Start by rewriting your parabola equation in the form \(x^2 = 4py\).
• Compare the coefficient of \(y\) in your equation to \(4p\) to solve for \(p\).
• Once you find \(p\), plug it into the focus formula \(0, p\).

In our exercise, \(p\) was found to be 2, so the focus is at \( (0, 2)\). This point is directly above the vertex when the parabola opens upwards. The focus is crucial because it helps to define the "path" of the parabola. Every point on the parabola is equidistant from the focus and a line known as the directrix.

Directrix of a Parabola

The directrix is an important line that is associated with a parabola. While the curve itself is defined by both a focus and this directrix, they work together to give the parabola its symmetric shape.

The directrix is a straight line that runs parallel to the parabola's axis of symmetry. For a vertical parabola in the form of \(x^2 = 4py\), the equation of the directrix is \(y = -p\). This means that once you determine the \(p\) from your equation, you can easily find the directrix.

• Identify \(p\) from the equation \(x^2 = 4py\).
• The directrix is located at \(y = -p\), just use the negative value of \(p\) you calculated.

In our parabola example, where \(p = 2\), the directrix lies at \(y = -2\). This line is found below the vertex at \(y = 0\), illustrating that the directrix exists opposite to the focus when the parabola opens upwards. Understanding the position of the directrix helps in reading and sketching the parabola accurately.

Focal Diameter of a Parabola

The focal diameter of a parabola is a specific measurement that helps describe its size and openness. Also known as the "latus rectum," this is a line segment that passes through the focus. It's perpendicular to the axis of symmetry of the parabola.

To calculate the focal diameter, you can simply use the formula \(\lvert 4p\rvert\). This straightforward calculation comes directly from having the parabola's equation in its standard form \(x^2 = 4py\).

• Extract the value of \(p\) similarly as before.
• Use the formula \(\lvert 4p\rvert\) to obtain the focal diameter.

For our parabola, with \(p = 2\), the focal diameter is \(\lvert 4 \times 2\rvert = 8\). This indicates that the line segment through the focus has a length of 8 units. A larger focal diameter means the parabola is wider, while a smaller focal diameter implies a narrower shape. Understanding the focal diameter provides insight into the overall geometry of the parabola.