Problem 87 - A Graphical Approach to Precalculus with Limits

Question

Find an equation of a parabola that satisfies the given conditions. Focus \((0,0) ;\) directrix \(x=-2\)

Step-by-Step Solution

Answer

The equation is \(y^2 = 4(x + 1)\).

Step 1: Understand the Parabola Definition

A parabola is the set of all points that are equidistant from a point (the focus) and a line (the directrix). For this problem, the focus is (0, 0) and the directrix is the line \(x = -2\).

Step 2: Identify the Parabola Form

Since the directrix is a vertical line and the focus is at (0, 0), the parabola opens horizontally. The equation will be of the form \((y-k)^2 = 4p(x-h)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus.

Step 3: Determine the Vertex's Coordinates

For a horizontal parabola, the vertex \((h, k)\) is midway between the focus and directrix. The x-coordinate \(h\) of the vertex is the average of the focus's x-coordinate and the directrix's x-value: \(h = \frac{0 + (-2)}{2} = -1\). The y-coordinate \(k\) is 0 since the focus and directrix are vertically aligned at y=0. Hence, the vertex is \((-1, 0)\).

Step 4: Find the Value of \(p\)

The value of \(p\) is the distance between the vertex and the focus. Because the vertex is at \((-1, 0)\) and the focus is at \((0, 0)\), \(p = 0 - (-1) = 1\).

Step 5: Construct the Parabola Equation

Substitute the values \(h = -1\), \(k = 0\), and \(p = 1\) into the parabola equation form: \((y - 0)^2 = 4(1)(x + 1)\). Simplifying, the equation becomes \(y^2 = 4(x + 1)\).

Key Concepts

Focus and Directrix

Understanding the focus and directrix is essential when it comes to working with parabolas. A parabola is defined as the set of points equidistant from a fixed point, called the focus, and a fixed line, known as the directrix.
For example, let's consider the given parabola where the focus is (0, 0) and the directrix is the line x = -2.

• The focus is the point where all the reflected lines from the parabola converge or appear to converge.
• The directrix is a line used as a reference to ensure the parabola is equidistant from the focus.

To verify if a point, say (x, y), lies on the parabola, one should check if the distance from (x, y) to the focus is the same as the perpendicular distance from (x, y) to the directrix.

Vertex of a Parabola

The vertex of a parabola, denoted as (h, k), is a pivotal point that lies exactly halfway between the focus and the directrix.
In our example, the focus is (0, 0), and the directrix is x = -2.

Finding the Vertex

The x-coordinate of the vertex is the average of the x-coordinates of the focus and directrix: \[ h = \frac{0 + (-2)}{2} = -1\]The y-coordinate k comes directly from the focus's and directrix's alignment along the y-plane, which is 0 here, giving the vertex coordinates as (-1, 0).

Importance of the Vertex

• The vertex can either be the lowest or highest point on the parabola, depending on its orientation.
• For a horizontally oriented parabola, like in this case, the vertex is neither the highest nor lowest, but it is a critical turning point.

Horizontal Parabola

A horizontal parabola opens sideways, either to the left or right. The orientation of the parabola depends on the direction from the focus to the vertex and the positioning of the directrix.
For the given problem, since the directrix x = -2 is vertical, and the focus is (0, 0), the parabola opens horizontally.

Equation of a Horizontal Parabola

The standard form for a horizontal parabola is:\[(y - k)^2 = 4p(x - h)\]Where:

• (h, k) is the vertex of the parabola. In our problem, it is (-1, 0).
• p is the distance from the vertex to the focus, which is 1 in this case.

Substituting these into the formula gives us:\[y^2 = 4(x + 1)\]

Characteristics of Horizontal Parabolas

• They do not have a minimum or maximum like vertical parabolas.
• They may intercept the y-axis depending on their position, although this one doesn't as due to (y - k)^2, y = 0 at the x-intercept.