Problem 18 - Precalculus Mathematics for Calculus

Question

Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$x=\frac{1}{2} y^{2}$$

Step-by-Step Solution

Answer

Focus: \((2,0)\); Directrix: \(x = -2\); Focal diameter: 4.

Step 1: Identify the Parabola's Orientation and Form

The given equation is \(x=\frac{1}{2}y^2\). This is in the form \(x = a(y-k)^2 + h\), which indicates a parabola that opens sideways. In this case, it opens to the right because \(a\), which is positive, is \(\frac{1}{2}\). Here \(h=0\) and \(k=0\).

Step 2: Find the Value of 'a'

The coefficient \(a\) in the equation \(x = a(y-k)^2 + h\) is the value that determines the direction and the width of the parabola. For the given parabola, \(a = \frac{1}{2}\).

Step 3: Determine the Focus

For a sideways parabola \((x = a(y-k)^2 + h)\), the distance from the vertex \((h, k)\) to the focus is \(\frac{1}{4a}\). Here, \(a = \frac{1}{2}\), so the distance is \(\frac{1}{4(\frac{1}{2})} = 2\). Since the parabola opens to the right, the focus is 2 units to the right of the vertex (0,0), so the focus is at \((2, 0)\).

Step 4: Determine the Directrix

The directrix of a parabola that opens right is a vertical line found \(2\) units to the left of the vertex (opposite the direction of the focus). Since our vertex is \((0,0)\), the directrix is the vertical line at \(x = -2\).

Step 5: Calculate the Focal Diameter

The focal diameter of a parabola is the absolute value of \(\frac{1}{a}\). Here \(a = \frac{1}{2}\), so the focal diameter is \(\frac{1}{\frac{1}{2}} = 2\times1 = 4\).

Step 6: Sketch the Graph

Draw the coordinate plane and plot the vertex at \((0,0)\). Locate the focus at \((2,0)\) and draw the directrix line \(x = -2\). Sketch the parabola opening to the right from the vertex, passing through points that maintain equal distance between the focus and the directrix, ensuring the graph is symmetric relative to the \(x\)-axis.

Key Concepts

Focus and Directrix

The focus and directrix are fundamental elements of a parabola. They help define the shape and position of the parabola on a graph.

• The focus is a point inside the parabola that, along with the directrix, helps shape the curve. For the given parabola, which opens to the right, the focus is located at (2, 0). This means it is 2 units to the right of the vertex (0,0).


• The directrix serves as a line that is the same distance away from the vertex as the focus is, but in the opposite direction. For this parabola, the directrix is a vertical line located at x = -2.

The key relationship in a parabola is that any point on it is equidistant from the focus and the directrix. This property is crucial for sketching and understanding the curve's shape.

Focal Diameter

The focal diameter of a parabola tells us the width of the parabola at the level of the focus. It is also known as the latus rectum. Understanding the focal diameter is essential for sketching and interpreting the parabola's geometry.

• The focal diameter is expressed as the absolute value of \( \frac{1}{a} \). For our parabola, where \( a = \frac{1}{2} \), the focal diameter is \( \frac{1}{\frac{1}{2}} = 4 \).


• This means that, at the level of the focus, the width of the parabola from one side to the other is 4 units.

Knowing the focal diameter helps us identify how "spread out" the parabola is, especially how wide it is at the focal level. This parameter can often be useful for practical applications, like determining the beam width in satellite dishes or the reflective properties of headlights.

Parabola Orientation and Form

Determining the orientation and form of a parabola is essential for understanding its full structure. Let's break these concepts down:

• The orientation refers to the direction in which the parabola opens. If the parabola's equation is in the form \( x = a(y-k)^2 + h \), it opens sideways. In our specific case, it opens to the right because the value of \( a = \frac{1}{2} \) is positive.


• The form can be given as the equation itself, such as \( x = \frac{1}{2}y^2 \). This helps to normalize the orientation with relation to the axis. Here, it makes clear the parabola is centered at the origin \((0,0)\) without any vertical or horizontal shifts \((h = 0, k = 0)\).

Both orientation and form help us not only plot the parabola accurately but also understand how changes in the coefficients \( h, k, \) and \( a \) might affect its position and width. Recognizing these changes over variations in the equation is key in learning how to manipulate and interpret parabolic graphs in mathematics.