Problem 22 - 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. $$x+\frac{1}{5} y^{2}=0$$

Step-by-Step Solution

Answer

Focus: \\(-\frac{5}{4}, 0\\), Directrix: \\(x = \frac{5}{4}\\), Focal Diameter: 5.

Step 1: Rewrite the Equation

To make it easier to identify the components of the parabola, let's rewrite the equation in the form of \(y^2 = 4px\).Starting from the equation \(x + \frac{1}{5}y^2 = 0\):\[x = -\frac{1}{5}y^2\]Multiply both sides by -5 to rearrange it:\[y^2 = -5x\]This equation is now in the standard form \(y^2 = 4px\) with \(4p = -5\).

Step 2: Find the Value of p

From the equation \(y^2 = 4px\), we know that \(4p = -5\). Solve for \(p\) by dividing both sides by 4:\[p = \frac{-5}{4}\]So, \(p = -\frac{5}{4}\).

Step 3: Find the Focus

The focus of a parabola \(y^2 = 4px\) is located at \(p, 0\). Using \(p = -\frac{5}{4}\), the focus is:\[\left(-\frac{5}{4}, 0\right)\]

Step 4: Determine the Directrix

The directrix of the parabola is a vertical line given by \(x = -p\). Substitute \(p = -\frac{5}{4}\):\[x = -(-\frac{5}{4}) = \frac{5}{4}\]Therefore, the directrix is the line \(x = \frac{5}{4}\).

Step 5: Calculate the Focal Diameter

The focal diameter of a parabola in the form \(y^2 = 4px\) is given by the absolute value of \(4p\). We already know \(4p = -5\), so:\[\text{Focal Diameter} = |4p| = 5\]

Step 6: Sketch the Parabola and Directrix

Draw the parabola \(y^2 = -5x\) with its vertex at the origin \(0,0)\). This parabola opens to the left. Plot the focus at \((-\frac{5}{4}, 0)\) and the directrix as the vertical line \(x = \frac{5}{4}\). Include the focal diameter which is the distance across the parabola through the focus.

Key Concepts

Focus of a Parabola

The focus of a parabola is a crucial point that helps define the shape and position of the parabola. For the given equation, let's start by identifying the focus of the parabola. When a parabola is expressed in the form \(y^2 = 4px\), the focus is located at \((p, 0)\). In our example, the equation was rearranged to \(y^2 = -5x\), which means \(4p = -5\). Solving for \(p\), we get \(p = -\frac{5}{4}\).
Therefore, the focus of this parabola is at the point \(\left(-\frac{5}{4}, 0\right)\). This point is essential in determining the direction the parabola opens and how it is graphed.

Directrix of a Parabola

The directrix is an imaginary line that, together with the focus, defines the parabola. For the equation \(y^2 = 4px\), the directrix is a vertical line given by \(x = -p\).
Using the previously calculated value of \(p = -\frac{5}{4}\), the equation for the directrix becomes \(x = \frac{5}{4}\).
This tells us that the directrix is a vertical line 1.25 units to the right of the y-axis. This line helps us understand how the parabola relates spatially to the focus and is parallel to the axis of symmetry of the parabola.

Focal Diameter

The focal diameter, also known as the latus rectum, is an important component that tells us about the "width" of the parabola at the focus. It is the distance across the parabola through the focus and parallel to the directrix.
For the standard form \(y^2 = 4px\), the focal diameter is given by \(|4p|\).
In this case, we found \(4p = -5\), so the focal diameter is \(|-5| = 5\).
Knowing the focal diameter is crucial for determining how steep or wide the parabola appears, which helps in accurately sketching or visualizing the curve.

Standard Form of a Parabola

Understanding the standard form of a parabola is essential to identify its key characteristics like focus, direction, and graphing.
The standard form of a parabola can be \(y^2 = 4px\) or \(x^2 = 4py\), which indicates parabolas that open horizontally or vertically, respectively.
In our problem, the equation \(y^2 = -5x\) indicates a horizontally opening parabola, specifically to the left.
Being familiar with this form allows quick identification of properties such as the position of the focus, the direction the parabola opens, and the equation of the directrix. This form organizes the essential components of the parabola for easy analysis and graphing.

Graphing Parabolas

Graphing parabolas involves not only plotting the curve but also understanding its structure through the focus, directrix, and vertex.
To graph the equation \(y^2 = -5x\), note the vertex at the origin \((0,0)\), focus at \(\left(-\frac{5}{4}, 0\right)\), and the directrix line \(x = \frac{5}{4}\).
This specific parabola opens to the left. Understanding that allows you to draw it accordingly.

• Start by marking the vertex and focus on a graph.
• Draw the directrix line as a guide.
• With these points plotted, sketch the curve maintaining symmetry about the focus and directrix.

This visualization aids students in comprehending the spatial relationship between the components of the parabola and the parabola itself.