Problem 17 - 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: \(\frac{1}{2}\). The parabola opens leftwards.

Step 1: Identify the form of the parabola equation

Comparing the given equation \( x = -2y^2 \) with the standard form of a parabola \( x = ay^2 \), we see that \( a = -2 \). Since \( a < 0 \), this parabola opens leftward.

Step 2: Find the vertex of the parabola

The standard form \( x = ay^2 \) represents a parabola with its vertex at the origin \((0, 0)\). This applies to our equation \( x = -2y^2 \).

Step 3: Determine the focus

The focus of a parabola \( x = ay^2 \) is located at \( (\frac{1}{4a} + h, k) \), where \((h, k)\) is the vertex. Substituting \( a = -2 \):\[\frac{1}{4(-2)} = -\frac{1}{8}\].Thus, the focus is at \((-\frac{1}{8}, 0)\).

Step 4: Find the directrix

The directrix of a parabola \( x = ay^2 \) is a vertical line located at \( x = h - \frac{1}{4a} \). Using \( a = -2 \):\[ x = 0 + \frac{1}{8} = \frac{1}{8} \].So, the directrix is the line \( x = \frac{1}{8} \).

Step 5: Calculate the focal diameter

The focal diameter is the absolute value of \( \frac{1}{a} \). With \( a = -2 \),\[ \frac{1}{-2} = -\frac{1}{2} \]. Thus, the focal diameter is \( \frac{1}{2} \).

Step 6: Sketch the graph

To sketch the graph, plot the vertex \( (0,0) \), the focus \( (-\frac{1}{8}, 0) \) on the left of the vertex, and draw the directrix as the line \( x = \frac{1}{8} \). The parabola should open towards the left, with its width determined by the focal diameter.

Key Concepts

Focus of a Parabola

The focus of a parabola is a crucial point that helps define the shape and direction of the parabola. In our exercise, the parabola given is in the form of \( x = -2y^2 \). The vertex is located at the origin \((0, 0)\). When identifying the focus for a parabola of the form \( x = ay^2 \), it is found at \( \left( \frac{1}{4a} + h, k \right) \), where \((h, k)\) is the vertex.
For this specific parabola:

• \( a = -2 \)
• Focus calculation: \( \frac{1}{4(-2)} = -\frac{1}{8} \)

Thus, the focus is at \((-\frac{1}{8}, 0)\), meaning it is situated to the left of the vertex. This point gives us one aspect of how the parabola is altered along the horizontal direction.

Directrix

The directrix complements the focus to form the set of guiding elements of a parabola. It is a line that helps determine the exact curvature of the parabola along with the focus. The equation for the directrix of a parabola in the form \( x = ay^2 \) is given by \( x = h - \frac{1}{4a} \), where \((h, k)\) is the vertex.
In our example:

• Directrix calculation: \( x = 0 - \left(-\frac{1}{8}\right) = \frac{1}{8} \)

Therefore, the directrix is the vertical line \( x = \frac{1}{8} \). It lies to the right of the vertex, offering a balance to the focus on the left. The parabola opens leftwards away from the directrix, indicating an inverse relationship between its location and the parabola's opening direction.

Focal Diameter

The focal diameter of a parabola is an important measurement representing the width of the parabola at the level of the focus. It is defined as the absolute value of the reciprocal of \( a \). This signifies how spread out or narrow the parabola is at its widest point, intersecting at the focus.
For the given equation \( x = -2y^2 \), calculate the focal diameter as follows:

• \( a = -2 \)
• Focal Diameter calculation: \( \left| \frac{1}{a} \right| = \left| \frac{1}{-2} \right| = \frac{1}{2} \)

The focal diameter, being \( \frac{1}{2} \), indicates this parabola is relatively narrow. The smaller focal diameter means the parabola does not spread widely as it shifts away from the vertex, especially since it opens horizontally.