Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 19-30: Find an equation of the parabola that satisfies the given conditions. Focus \(F(6,4)\) $$ \text { directrix } y=-2 $$

Step-by-Step Solution

Answer

The equation of the parabola is \((x-6)^2 = 12(y-1)\).

Step 1: Identify the standard form

The equation of a parabola with a vertical axis of symmetry can be written as \((x-h)^2 = 4p(y-k)\), where \((h,k)\) is the vertex and \(p\) is the distance from the vertex to the focus or directrix.

Step 2: Determine the vertex

The vertex \((h, k)\) is the midpoint between the focus and the directrix. Here, the focus is \((6,4)\) and the directrix is \(y = -2\). The y-coordinate of the vertex is \(k = \frac{4 + (-2)}{2} = 1\). Since the axis of symmetry is vertical and by looking at the x-coordinate of the focus, we have the vertex \((6,1)\).

Step 3: Calculate the distance \(p\)

\(p\) is the distance from the vertex to the focus or directrix. From vertex \((6,1)\) to focus \((6,4)\), the distance \(p\) is \(|4-1| = 3\).

Step 4: Write the equation of the parabola

Substitute \(h = 6\), \(k = 1\), and \(p = 3\) into the standard form \((x-h)^2 = 4p(y-k)\). Thus, we have \((x-6)^2 = 4(3)(y-1)\) or \((x-6)^2 = 12(y-1)\).

Key Concepts

Focus of a Parabola

The focus of a parabola is a single point located inside the curve and plays a crucial role in its geometric properties. This point determines how the parabola opens and how it relates to its directrix. The distances from any point on the parabola to the focus and the directrix are always equal. In our example, the focus is given as the point \((6,4)\). This means every point on the parabola is equidistant from this focus and the directrix, \(y = -2\). The focus helps in defining the curve's path, making it an essential part of the parabola's structure.

Directrix of a Parabola

The directrix of a parabola is a fixed line that helps in defining and shaping the curve. It is not intersected by the parabola itself, rather it serves as a reference line for measuring distances. For a vertical parabola, the directrix is horizontal. In the exercise given, the directrix is \(y = -2\). All points on the parabola are equidistant from this line and the focus. This relationship aids in determining the vertex and the overall equation of the parabola.

Vertex of a Parabola

The vertex of a parabola is the point that lies exactly halfway between the focus and the directrix. It is the parabola's turning point and can either represent a minimum or maximum of the curve, depending on its orientation. To find the vertex, you average the y-coordinates of the focus and directrix in a vertical parabola. In our problem, the focus at \((6,4)\) and directrix \(y = -2\) give the vertex at \((6,1)\). The vertex is crucial as it is used in the parabola’s equation as the point \((h, k)\).

Distance Formula in Geometry

The distance formula in geometry provides a way to calculate the distance between two points in a plane. For two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance is calculated as:

• Distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

In the problem, calculating the distance \(p\) from the vertex \((6,1)\) to the focus \((6,4)\) yields \(|4 - 1| = 3\). This number \(p\) is essential for writing the equation in standard form.

Standard Form of a Parabola

The standard form of a parabola provides a structured way to express its equation. For a vertical parabola, this form is

• \((x-h)^2 = 4p(y-k)\)

Here, \((h, k)\) represents the vertex, and \(p\) is the distance from the vertex to the focus. In this exercise, substituting \(h = 6\), \(k = 1\), and \(p = 3\) into the equation results in

• \((x-6)^2 = 12(y-1)\)

This equation describes the parabola completely, showing how all points relate to its vertex, focus, and directrix.