Problem 5 - Algebra 2

Question

Write an equation for a graph that is the set of all points in the plane that are equidistant from the given point and the given line. $$ F(0,4), y=0 $$

Step-by-Step Solution

Answer

The equation of the graph that is the set of all points in the plane that are equidistant from the point \(F(0, 4)\) and the line \(y = 0\) is \(y = -1/4x^2 + 4\).

Step 1: Identify the Vertex and Directrix

From the exercise problem, we are given that the vertex is \(F(0,4)\) and the directrix is \(y = 0\). Therefore, we have our vertex as \((h, k) = (0, 4)\).

Step 2: Find the Distance between the Vertex and Directrix

The vertical distance from the vertex to the directrix is \(4 - 0 = 4\) units.

Step 3: Construct the Equation for the Parabola

Since the parabola opens downwards and the distance from the vertex to the directrix is \(4\), the value of \(a\) in the equation \(y = a(x - h)^2 + k\) should be \(-1/4\) to make the points equidistant. Hence, the final equation for the parabola would be \(y = -1/4x^2 + 4\).

Key Concepts

Understanding the Vertex and Directrix

In the context of parabolas, the **vertex** and **directrix** play crucial roles in determining its shape and position. The vertex is the point located at the tip of the parabola, and it marks the point closest to the directrix. In our problem, the vertex is given as \( F(0,4) \). This means we locate the vertex at the point \( (0, 4) \) in the coordinate plane.

The **directrix** is essentially a line that serves as a fixed reference from which distances are measured to determine the parabola's points. For any point on the parabola, the distance to the directrix is equal to the distance to the focus. In our problem, the directrix is the horizontal line \( y = 0 \). Identifying these components helps us construct the parabola equation by ensuring correct measurement of distances.

What are Equidistant Points?

The concept of **equidistant points** in parabolas refers to the property that points on the curve are always at equal distances from the vertex and the directrix. Let’s delve into what this means.

Every point \( P \) on the parabola satisfies a unique condition: the distance from \( P \) to the vertex (which is also called the focus in this scenario) is equal to the distance from \( P \) to a line called the directrix. This balance is a defining property of parabolas that guides their distinct shape.

• This means that if you draw a straight line from any point \( P \) on the parabola to the focus and another straight line from the same \( P \) to the directrix, these lines will always have the same length.
• The parabola ensures symmetry and precision in being equidistant around its shape.

Understanding this property is key to working with and visualizing how parabolas function in mathematics.

Understanding Parabola Orientation

A parabola’s **orientation** describes the direction in which it opens. This is determined by its equation's leading coefficient. Let's consider how orientation is set in our problem.

The equation formulated is \( y = -\frac{1}{4}x^2 + 4 \), which gives us useful orientation information:

• The negative sign before \( \frac{1}{4} \) indicates that the parabola opens downwards.
• If the coefficient were positive, it would open upwards instead.
• Parabolas may also open left or right if the equation swaps the \( x \) and \( y \) variables.

Orientation impacts how we graph a parabola and predict its behavior in solving practical problems. Coupling this understanding with vertex and directrix locations can create an accurate graphical representation.

Calculating Distance in Coordinate Geometry

**Distance in coordinate geometry** allows us to measure the separation between points or a point and a line, like the directrix, directly on the coordinate plane. It’s a fundamental skill that enables the calculation of distances in various geometric contexts.

In this exercise, calculating the distance involves:

• Identifying the coordinates of the vertex from the equation, \( (h, k) \).
• Determining the reference line, here, \( y = 0 \), and calculating the perpendicular distance from the vertex to this line is easy, since the vertical distance from \( y = 4 \) to \( y = 0 \) is simply \( 4 \) units.
• This distance affects the formulation of the parabola's equation by helping determine its vertical stretch and placement.

Mastering distance calculations in geometry is essential for solving complex problems involving shapes and graphs over distances.