Problem 13 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the vertex, focus, and directrix of the parabola. (b) Sketch a graph showing the parabola and its directrix. $$(x-3)^{2}=8(y+1)$$

Step-by-Step Solution

Answer

Vertex: (3, -1), Focus: (3, 1), Directrix: y = -3; Upward-opening parabola.

Step 1: Identify the Form of the Equation

The equation given is \((x-3)^2 = 8(y+1)\). This equation is in the form \((x-h)^2 = 4p(y-k)\), which represents a parabola that opens upwards or downwards.

Step 2: Determine Vertex Coordinates

For the standard form \((x-h)^2 = 4p(y-k)\), the vertex is \((h, k)\). Comparing this with the given equation \((x-3)^2 = 8(y+1)\), we see that \(h = 3\) and \(k = -1\). Thus, the vertex is \((3, -1)\).

Step 3: Find the Value of \(p\)

In the equation \((x-h)^2 = 4p(y-k)\), \(4p\) is the coefficient of the \((y-k)\) term. Here, \(4p = 8\), so \(p = 2\). The value of \(p\) determines the distance from the vertex to the focus and the directrix.

Step 4: Determine the Focus

The focus of a parabola that opens upwards is \((h, k + p)\). Since \(h = 3\), \(k = -1\), and \(p = 2\), the focus is \((3, -1+2)\) which simplifies to \((3, 1)\).

Step 5: Determine the Directrix

The directrix of a parabola that opens upwards is \(y = k - p\). With \(k = -1\) and \(p = 2\), the directrix is \(y = -1-2\), which simplifies to \(y = -3\).

Step 6: Sketch the Parabola

To sketch the parabola, plot the vertex at \((3, -1)\), the focus at \((3, 1)\), and draw the line \(y = -3\) as the directrix. Since the parabola opens upwards (as indicated by \((x-h)^2\)), sketch the curve such that it passes through the vertex and moves up away from the directrix.

Key Concepts

Vertex

The vertex of a parabola is the point where the parabola changes direction. This is a critical point that determines the general shape and position of the parabola on a graph. In the given parabola equation \((x-3)^2 = 8(y+1)\), we can determine the vertex by observing the equation's format.

The format \((x-h)^2 = 4p(y-k)\) reveals that the vertex \((h, k)\) is at \((3, -1)\). Here:

• \(h = 3\): indicating a horizontal displacement to the right from the origin.
• \(k = -1\): indicating a vertical shift downwards from the origin.

Thus, the vertex \(3, -1\) serves as a central point for sketching and helps in identifying the focus and directrix.

Focus

The focus of a parabola is a crucial concept that works together with the directrix to define the set of points that form the parabola. In simpler terms, the focus is a point inside the parabola that is used to define its curve.

For our equation \((x-3)^2 = 8(y+1)\), which opens upwards:

• The focus is calculated using the coordinates of the vertex \(h, k\) and the parameter \p\.
• Given that \p = 2\, the focus is located at \(h, k + p\) which simplifies to \(3, -1 + 2\) or \(3, 1\)\.

The focus \(3, 1\) offers the same distance from the vertex as the directrix, but in the upward direction, confirming the parabola's upward opening.

Directrix

The directrix of a parabola is a line that is used along with the focus to derive the shape of the parabola. Each point on the parabola is equidistant from the directrix and the focus, creating the parabolic shape.

For the equation \((x-3)^2 = 8(y+1)\):

• The directrix is always positioned a distance \p\ away from the vertex.
• Since \p = 2\ and the parabola opens upwards, the directrix can be calculated as \y = k - p\, which gives us \y = -1 - 2\ or \y = -3\.

This line \(y = -3\) sits below the vertex, balancing the parabola along with the focus, thus maintaining its precise symmetrical nature.