Problem 8 - Precalculus: Mathematical for Calculus

Question

Find the vertex, focus, and directrix of the parabola, and sketch the graph. $$y^{2}=16 x-8$$

Step-by-Step Solution

Answer

Vertex: \((\frac{1}{2}, 0)\); Focus: \((\frac{9}{2}, 0)\); Directrix: \(x = -\frac{7}{2}\).

Step 1: Understand the Parabola Equation

The given equation is \( y^2 = 16x - 8 \), which can be rewritten in a form that highlights its structure. This equation is of the form \( y^2 = 4px \), where \( p \) is the parameter that helps in identifying the focus and directrix. However, we'll need to rewrite this equation first.

Step 2: Rewrite the Equation in Standard Form

Rewrite the equation by moving the constant to the other side: \( y^2 = 16x - 8 \rightarrow y^2 = 16(x - \frac{1}{2}) \). This matches the standard form \( y^2 = 4p(x - h) \) with \( h = \frac{1}{2} \) and \( 4p = 16 \). Thus, \( p = 4 \).

Step 3: Find the Vertex

The vertex form \( y^2 = 4p(x - h) \) reveals that the vertex is at \((h, k)\), where \( k = 0 \) here. Hence, the vertex is at \( (\frac{1}{2}, 0) \).

Step 4: Determine the Focus

For a parabola in the form \( y^2 = 4p(x - h) \), the focus is at \((h + p, k)\). Using \( h = \frac{1}{2} \) and \( p = 4 \), the focus is located at \((\frac{1}{2} + 4, 0) = (\frac{9}{2}, 0)\).

Step 5: Find the Directrix

The directrix of the parabola \( y^2 = 4p(x - h) \) has the equation \( x = h - p \). Therefore, using \( h = \frac{1}{2} \) and \( p = 4 \), the directrix is \( x = \frac{1}{2} - 4 = -\frac{7}{2} \).

Step 6: Sketch the Graph

The parabola opens towards the right because the \( y^2 \) term is positive and it matches the opening towards \( x \). Plot the vertex at \( (\frac{1}{2}, 0) \), the focus at \((\frac{9}{2}, 0)\), and draw a vertical line at \( x = -\frac{7}{2} \) for the directrix. The parabola should be symmetric about the x-axis and have its arms opening towards the positive x-direction.

Key Concepts

Vertex

In a parabola, the vertex is a crucial point where the curve turns. For the equation given as \[ y^2 = 16x - 8 \]we need to rewrite it in the standard form \[ y^2 = 4p(x - h) \] to find the vertex. By rearranging the equation, we obtain \[ y^2 = 16(x - \frac{1}{2}) \], indicating that the vertex is at \((h, k)\).

• Here, \( h = \frac{1}{2} \) and \( k = 0 \).
• Therefore, the vertex is located at \( (\frac{1}{2}, 0) \).

The vertex provides a key reference point for sketching the parabola. It is the turning point from which the parabola opens right, left, up, or down. In this case, since the equation is derived in terms of \( x \), the opening direction is determined by how \( y^2 \) relates to \( x \). The vertex here shows the starting point of this rightward opening curve.

Focus

The focus of a parabola is a special point that illustrates the property of all rays entering a parabolic mirror converging at this exact spot. In the equation \[ y^2 = 4p(x - h) \],the focus is calculated using the parameters \( h \) and \( p \). For our parabola:

• We determined \( p = 4 \) and \( h = \frac{1}{2} \).
• The formula for the focus is \( (h+p, k) \).
• Substituting the values, the focus is at \((\frac{9}{2}, 0)\).

Understanding where the focus is helps us comprehend the parabolic shape’s reflective properties. This is particularly useful in optics and acoustics, where the idea of a focus is translated into gathering or dispersing energy effectively at a point.

Directrix

The directrix is a unique line associated with a parabola that serves as a geometric reference. The position of the directrix provides insight into the balancing forces that shape the parabola.For our equation in the form \[ y^2 = 4p(x - h) \],the directrix has the equation \[ x = h - p \]. Here's how it's calculated for our parabola:

• With known constants \( h = \frac{1}{2} \) and \( p = 4 \),
• The directrix is given by the equation \( x = -\frac{7}{2} \).

The directrix aids in the construction of the parabola as it represents a line equidistant in a reflective property alongside the focus from any point on the parabola. It's on the opposite side of the focus, and its position helps to emphasize the open nature of the curve towards the right, as observed in our instance.