Problem 51 - Algebra and Trigonometry

Question

(a) Find equations for the family of parabolas with vertex at the origin and with directrixes \(y=\frac{1}{2}, y=1, y=4,\) and \(y=8\) . (b) Draw the graphs. What do you conclude?

Step-by-Step Solution

Answer

The equations are \( y = 2x^2 \), \( y = x^2 \), \( y = \frac{1}{4}x^2 \), and \( y = \frac{1}{16}x^2 \). As the directrix moves away, the parabolas widen.

Step 1: Understanding the Problem

We need to find equations for parabolas with vertex at the origin and different directrices: \( y = \frac{1}{2} \), \( y = 1 \), \( y = 4 \), and \( y = 8 \). We will use the formula for a parabola with vertical axis of symmetry and its vertex at the origin, given the directrix \( y = k \): \[ y = \frac{1}{4a}x^2 \] where \( a \) is the distance from the vertex to the focus, and also to the directrix.

Step 2: Equation with Directrix at y = 0.5

For the directrix \( y = \frac{1}{2} \), the focus is at \( y = -\frac{1}{2} \), thus \( a = \frac{1}{4} \). This gives us the equation: \( y = 2x^2 \).

Step 3: Equation with Directrix at y = 1

For the directrix \( y = 1 \), the focus is at \( y = -1 \), so \( a = \frac{1}{2} \). The equation becomes \( y = x^2 \).

Step 4: Equation with Directrix at y = 4

For the directrix \( y = 4 \), the focus is at \( y = -4 \), so \( a = 2 \). The equation then is \( y = \frac{1}{4}x^2 \).

Step 5: Equation with Directrix at y = 8

For the directrix \( y = 8 \), the focus is at \( y = -8 \), therefore \( a = 4 \). The corresponding equation is \( y = \frac{1}{16}x^2 \).

Step 6: Graphing the Parabolas

To graph these parabolas, each has a vertex at the origin and opens upwards with varying degrees of "spread" or "narrowness," determined by the distance to the directrix. As \( y \) becomes larger, the parabola becomes wider.

Step 7: Conclusion from Graphs

From the graphs, we observe that as the directrix moves further away from the vertex (increasing \( y \)), the parabola becomes wider. This is because the focus, being equidistant from the vertex, moves further away, increasing the value of \( a \) and therefore decreasing the sharpness or "narrowness" of the parabola.

Key Concepts

Vertex

The vertex of a parabola is a critical point where the curve changes direction. In the exercises we examined, the vertex for all the parabolas is positioned at the origin, which is the point \(0, 0\).
This simplifies the equation for the parabola since it eliminates any horizontal or vertical shifts. Thus, at the origin, the equation \( y = \frac{1}{4a}x^2 \) can be directly applied.

Key characteristics of a vertex include:

• It serves as the midpoint between the focus and directrix.
• It is the "turning" point of the parabola, where the vertical symmetry axis passes through.

For these specific parabolas with their vertex at the origin, this point is crucial in ensuring the curve maintains its correct position and shape.

Directrix

The directrix is an invisible horizontal line that relates spatially to the parabola. In our exercise, different values were assigned to the directrix, namely \(y = \frac{1}{2}\), \(y = 1\), \(y = 4\), and \(y = 8\). This line plays a fundamental role in determining the curvature and width of a parabola.

The significance of the directrix lies in:

• Its position relative to the vertex; it defines one half of the distance, \(a\), used in the parabola's equation.
• Influencing the "spread" of the parabola. A closer directrix results in a narrower curve, while a further directrix leads to a wider parabola.

In essence, as the directrix moves up (y increasing), the parabola becomes wider due to the increasing "a," thus affecting how the parabola is shaped and how it opens.

Focus

The focus is a crucial point inside the parabola's curve. It is paired with the directrix to define the properties of the parabola. In our exercises, the position of the focus can be deduced from the location of the directrix, considering the vertex is at the origin.

Here’s how the focus works with the directrix:

• The focus is located equidistant from the vertex as the directrix, but on the opposite side.
• Its calculated distance, \(a\), is vital for the parabola's equation, determining how the curve folds in or out.

For example, if the directrix is \(y = 4\), the focus is \(y = -4\), creating a balanced structure that defines the parabola's shape. Moving the focus alters this balance as it establishes the curve's depth and stretch.

Equations of Parabolas

The equation of a parabola gives a precise mathematical relationship describing its shape. In this problem set, we derived specific equations from given directrix values while keeping the vertex constant at the origin.

Using the general formula \( y = \frac{1}{4a}x^2 \), adjustments are made based on the directrix values, determining the parameter \(a\).

For instance, the equations derived include:

• For directrix \(y = \frac{1}{2}\): \( y = 2x^2 \)
• For directrix \(y = 1\): \( y = x^2 \)
• For directrix \(y = 4\): \( y = \frac{1}{4}x^2 \)
• For directrix \(y = 8\): \( y = \frac{1}{16}x^2 \)

These equations highlight how changing the directrix distance modifies the parabola's "a" value, affecting its width. The ability to interpret these equations helps in understanding how to manipulate the loci of points that form the shape of any given parabola.