Problem 21 - 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. $$x^{2}+12 y=0$$

Step-by-Step Solution

Answer

The focus is (0, -3), the directrix is y = 3, and the focal diameter is 12.

Step 1: Identify the Standard Form of the Parabola

The given equation is \(x^2 + 12y = 0\). A parabolic equation of the form \(x^2 = 4py\) is a vertical parabola opening upwards or downwards. By rearranging, \(x^2 = -12y\), which matches this form.

Step 2: Determine the Vertex and Orientation

The vertex form \(x^2 = 4py\) indicates that the vertex is at the origin \((0, 0)\). Since the coefficient of \(y\) is negative, the parabola opens downwards.

Step 3: Calculate 'p' and Determine the Focus

In \(x^2 = -12y\), the term \(-12\) is equal to \(4p\). Solving for \(p\), we have \(4p = -12\), so \(p = -3\). The focus, \((0, p)\), is \((0, -3)\).

Step 4: Identify the Directrix

The directrix for a parabola with vertex at the origin and opening vertically is \(y = -p\). Since \(p = -3\), the directrix is \(y = 3\).

Step 5: Calculate the Focal Diameter

The focal diameter, which is the length of the latus rectum, is \(|4p| = |4 \times -3| = 12\).

Step 6: Sketch the Parabola with the Focus and Directrix

Draw the coordinate plane. Plot the vertex at (0, 0). Mark the focus at (0, -3). Draw the directrix line \(y = 3\). Sketch a symmetric parabola opening downwards with respect to this axis.

Key Concepts

Focus of a Parabola

The focus of a parabola is a crucial point that helps define the curve's shape and position. If you imagine a flashlight's beam, the focus is like the bulb where all the light originates.
For a vertical parabola given in the standard form of \(x^2 = 4py\), the focus lies on the axis of symmetry. When the vertex is at the origin,

• The focus will be at (0, p) if the parabola opens upwards or downwards.
• Here, since our equation transformed into \(x^2 = -12y\), solving for \(p\) gives us \(p = -3\).
• Thus, the focus for this parabola is at (0, -3), meaning it is directly 3 units below the origin.

The position of the focus is vital because it's where the distances to the parabola are equal from any point on the curve to the directrix.

Directrix of a Parabola

The directrix is an imaginary line associated with the parabola that, together with the focus, helps define the curve. It acts as a guide for the shape of the parabola, and every point on the parabola is equidistant to both the focus and this line.
In our vertical parabola given by \(x^2 = -12y\):

• The formula for the directrix is \(y = -p\).
• With \(p = -3\), the directrix is \(y = 3\).

This line is perpendicular to the axis of symmetry and is located above the parabola opening downwards from the focus. Understanding the directrix is essential for students as it provides a straightforward method to sketch the geometric properties of the parabola.

Focal Diameter

The focal diameter, or latus rectum, of a parabola, is a line segment perpendicular to the axis of symmetry that passes through the focus and touches the parabola at two points. It provides an intuitive sense of the parabola's width at the focus.
Calculating the focal diameter involves the absolute value of \(4p\). For this scenario:

• With \(x^2 = -12y\), \(p = -3\).
• The focal diameter is \(|4p| = |4 \times -3| = 12\).

This length of 12 units indicates how 'wide' the parabola is at the level of the focus. Visualizing the focal diameter can be extremely helpful when sketching the parabola as it emphasizes the symmetric breadth of the curve at its focal point.