Problem 14 - 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. $$y^{2}=16 x$$

Step-by-Step Solution

Answer

Focus: (4, 0), Directrix: x = -4, Focal diameter: 16.

Step 1: Identify the Form of the Parabola

The given equation is \(y^2 = 16x\). This is in the form \(y^2 = 4px\), which indicates that the parabola opens rightward.

Step 2: Find the Value of 'p'

Since the equation \(y^2 = 4px\) is equivalent to \(y^2 = 16x\), we equate and solve for \(p\): \(4p = 16\). Thus, \(p = 4\).

Step 3: Determine the Focus

The focus of a parabola \(y^2 = 4px\) is at \((p, 0)\). Given that \(p = 4\), the focus is at \((4, 0)\).

Step 4: Determine the Directrix

The directrix of a parabola \(y^2 = 4px\) is a vertical line at \(x = -p\). Thus, the directrix is \(x = -4\).

Step 5: Calculate the Focal Diameter

The focal diameter is the absolute value of \(4p\). Substitute \(p = 4\): Focal diameter = \(|4 * 4| = 16\).

Step 6: Sketch the Parabola and Directrix

Draw the parabola opening to the right. The vertex is at the origin \((0,0)\), the focus at \((4,0)\), and the directrix is the vertical line \(x = -4\). The parabola passes through points \((4, 4)\) and \((4, -4)\) which helps to visualize it.

Key Concepts

Focus of Parabola

The focus of a parabola is a special point that helps define its shape and orientation. It is one of the most essential features to understand when dealing with parabolas. This point lies on the axis of symmetry, and every point on the parabola is equidistant from the focus and the directrix. For the quadratic equation given as \( y^2 = 4px \), the parabola opens rightwards if \( p > 0 \), as in our case.To find the focus, we use the formula for a horizontal parabola:

• The focus is located at \((p, 0)\).

For the equation \( y^2 = 16x \), we equate \( 16x \) to \( 4px \) giving us \( p = 4 \). Therefore, the focus of this parabola is positioned at \((4, 0)\).The focus is crucial in terms of reflection. Light or signals hitting the parabolic curve will converge or appear to emanate from the focus. This property makes parabolas indispensable in satellite dishes and telescope mirrors.

Directrix of Parabola

The directrix of a parabola is a line that, along with the focus, defines and governs the behavior of the parabola. It works together with the focus to maintain the parabola's symmetrical shape. Every point on the parabola maintains equal distance from both the focus and the directrix.For a parabola expressed by \( y^2 = 4px \), the directrix is described by a vertical line defined by:

• Equation of the directrix: \( x = -p \).

For the given equation \( y^2 = 16x \), we have already found \( p = 4 \). Hence, the directrix of this parabola is the line \( x = -4 \).The directrix can be thought of as a boundary that the parabola never crosses, effectively setting one side of the parabola's "measured balance" with its focus.

Focal Diameter

The focal diameter of a parabola is the length of the chord that passes through its focus and is perpendicular to the axis of symmetry. This term is sometimes also referred to as the "latus rectum" of the parabola.For a parabola defined by \( y^2 = 4px \), the focal diameter is calculated as the absolute value of \( 4p \). It provides insights into the width of the parabola. Greater focal diameters mean the parabola is "wider" at its focus point.In this exercise, given \( y^2 = 16x \) and solving for \( p \) yields \( p = 4 \), leading to:

• Focal diameter = \(|4 \times 4| = 16\).

This means the parabola, when intersected through its focus, has a span of 16 units, which helps visualize and sketch its opening width. Understanding this concept aids in precisely graphing and recognizing the parabola's breadth.