Problem 16 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$x=2 y^{2}$$

Step-by-Step Solution

Answer

Focus: \((\frac{1}{8}, 0)\), Directrix: \(x = -\frac{1}{8}\), Focal Diameter: \(\frac{1}{2}\).

Step 1: Identify the Parabola Type

The given equation is \( x = 2y^2 \). This is in the form \( x = ay^2 \). Parabolas in this form open horizontally, specifically to the right if \( a > 0 \) and to the left if \( a < 0 \).

Step 2: Convert to Standard Form

The standard form of a horizontally opening parabola is \( (y-k)^2 = 4p(x-h) \). Comparing \( y^2 = \frac{1}{2}x \) to this form, we identify \( h = 0 \), \( k = 0 \), and \( 4p = \frac{1}{2} \). So, \( p = \frac{1}{8} \).

Step 3: Find the Focus

The vertex of the parabola is at \((h, k) = (0, 0)\). The focus of the parabola is \((h+p, k)\), so it is at \(\left( \frac{1}{8}, 0 \right)\).

Step 4: Find the Directrix

The directrix is a vertical line given by \( x = h - p \). Substituting the known values gives \( x = 0 - \frac{1}{8} = -\frac{1}{8} \).

Step 5: Calculate the Focal Diameter

The focal diameter is the absolute value of \(4p\). From earlier calculation \( 4p = \frac{1}{2} \). So the focal diameter is \( \frac{1}{2} \).

Step 6: Sketch the Graph

On the graph, plot the vertex at \((0, 0)\), the focus at \(\left( \frac{1}{8}, 0 \right)\), and the directrix line at \( x = -\frac{1}{8} \). Sketch the parabola opening to the right, ensuring that it passes through the vertex and is symmetric about the x-axis.

Key Concepts

Focus of Parabola

The focus of a parabola is a crucial component that defines its shape and direction. Think of the focus as a special point where light or signals get directed or reflected, considering parabolas have reflective properties. For our horizontal parabola given by the equation \(x = 2y^2\), we've found that the vertex is located at the origin \((0,0)\). To determine the focus, we use the relation:

• The vertex form of the equation is \((y-k)^2 = 4p(x-h)\)
• Parameters: \( (h,k) = (0,0) \) and \(p = \frac{1}{8}\)

Thus, the focus can be found at \((h+p, k)\), which calculates to \(\left( \frac{1}{8}, 0 \right)\). The focus lies on the axis of symmetry of the parabola, confirming it is to the right of the vertex by \(\frac{1}{8}\) units.

Directrix of Parabola

A parabola's directrix is an invisible guiding line. It is critical as it works in tandem with the focus to define what points lie on the parabola. Every point on the parabola is equidistant from the focus and this line, which acts like a boundary on the other side. For the equation \(x = 2y^2\), the directrix is calculated by knowing:

• The equation \((y-k)^2 = 4p(x-h)\)
• We found \( (h,k) = (0,0) \) and \( p = \frac{1}{8} \)

The directrix is thus \(x = h - p\), which simplifies to \(x = -\frac{1}{8}\). It's positioned \(\frac{1}{8}\) units left of the vertex, marking a vertical line parallel to the parabola's path.

Focal Diameter

The focal diameter of a parabola, also known as the latus rectum, provides insight into the parabola's width. It's derived from the relationship between points on the curve and focal distances. In essence, it measures the distance across the parabola at the focus level. Calculate it using:

• The general form: \(4p\)
• Where previously identified \(p = \frac{1}{8}\)

Thus, the focal diameter is set as \(\frac{1}{2}\). This means if you draw a line segment across the parabola through the focus, it should measure \(\frac{1}{2}\) units. This metric helps in graphing the parabola symmetrically around the horizontal axis, ensuring precision in its curvature.