Problem 50 - College Algebra with Corequisite Support

Question

For the following exercises, find the equation of the parabola given information about its graph. Vertex is (1,2)\(; \quad\) directrix is \(y=\frac{11}{3},\) focus is \(\left(1, \frac{1}{3}\right)\).

Step-by-Step Solution

Answer

The equation of the parabola is \((x-1)^2 = -\frac{20}{3}(y-2)\).

Step 1: Identify Parabola Direction

Since the vertex is (1,2) and the focus is at \( \left(1, \frac{1}{3}\right) \), the parabola opens downwards because the focus is below the vertex.

Step 2: Use the Definition of Parabola

A parabola is the set of all points that are equidistant from a point (focus) and a line (directrix). Here, we use \( y = \frac{11}{3} \) as the directrix and the vertex as a reference to write the equation.

Step 3: Determine Vertex Form of the Parabola

The general vertex form of a parabola is \( (x-h)^2 = 4p(y-k) \), where \((h,k)\) is the vertex. The vertex is (1,2), thus: \((x-1)^2 = 4p(y-2)\).

Step 4: Find \(p\) Value

The distance \( p \) between the vertex and the directrix or focus is based on the formula \( (k - \text{focus y-coordinate}) = p \). Since the vertex y-coordinate \( k = 2 \) and the focus y-coordinate is \( \frac{1}{3} \), \( p = \frac{2 - \frac{1}{3}}{2} = \frac{5}{3} \).

Step 5: Write the Equation

Substitute \( p = -\frac{5}{3} \) (since it is downward opening) into the equation \( (x-1)^2 = 4p(y-2) \) to get \( (x-1)^2 = -\frac{20}{3}(y-2) \).

Key Concepts

Vertex Form

The vertex form of a parabola's equation is an incredibly useful tool for graphing and understanding parabolas. In vertex form, the equation is written as:

• \((x-h)^2 = 4p(y-k)\)

Here,

• \((h, k)\) represents the vertex of the parabola.
• \(p\) represents the distance from the vertex to the focus, as well as to the directrix. It determines how wide and in which direction the parabola opens.

For a parabola that opens vertically (either upwards or downwards), \((x-h)^2 = 4p(y-k)\) is used. This makes plotting and finding axes of symmetry straightforward. The vertex form is particularly beneficial because it gives us direct insight into the parabola's orientation and location on the graph. If you know the vertex and value of \(p\), constructing the equation is a breeze!

Parabola Direction

Determining the direction in which a parabola opens is crucial for correctly sketching its graph. The focus and vertex play a central role in this process. In many cases, you can quickly assess the direction by analyzing the positions of the vertex and focus relative to each other. For example:

• If the focus is above the vertex, the parabola opens upwards.
• If the focus is below the vertex, the parabola opens downwards.
• If the focus is to the left of the vertex, the parabola opens to the left.
• If the focus is to the right of the vertex, the parabola opens to the right.

In our exercise, the vertex is \((1, 2)\) and the focus is \(\left(1, \frac{1}{3}\right)\). Since the focus is below the vertex, we conclude the parabola opens downward. Understanding this helps when assembling the equation in vertex form and confirming the orientation matches given parameters.

Focus and Directrix

The focus and directrix are essential components in defining a parabola. A parabola can be considered as the set of all points equidistant from a fixed point, known as the focus, and a fixed line, the directrix. In this context:

• The focus, \(\left(1, \frac{1}{3}\right)\), is below the vertex, indicating that the parabola opens downwards.
• The directrix is the line \(y = \frac{11}{3}\), positioned above the vertex. This gives us a line of comparison to ensure all points are equidistant as required.

The position of the focus and the distance of points on the parabola from this line formulate the parabola's shape. When these distances are identical on all sides, the parabola retains a characteristic symmetric elegance.

Distance Formula

The distance formula is vital for calculating distances in coordinate geometry. In the context of parabolas, it aids in ensuring that each point on the parabola is equidistant from the focus and the directrix. Using the standard distance formula:

• For two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

This principle helps in situations where the precise distance must be computed to ensure the parabola's symmetry is correct. Once the distance from a point on the parabola to the focus matches the distance from the same point to the directrix, the shape of the parabola is secure. In our equation, this principle ensures that the vertex equation is consistent with all given characteristics.