Problem 6 - 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. $$ F\left(\frac{1}{2}, 0\right), x=-\frac{1}{2} $$

Step-by-Step Solution

Answer

The equation of the graph that is the set of all points equidistant from the point \(F(\frac{1}{2}, 0)\) and the line \(x=-\frac{1}{2}\) is \(y^2 = 2x\).

Step 1: Analyze the Given Information

Identify the focus \(F(\frac{1}{2}, 0)\) of the parabola and the line \(x=-\frac{1}{2}\) which serves as the directrix. The vertex of the parabola will be halfway between them, so on the line \(x = 0\). We also need to identify the orientation of the parabola. As the focus is to the right of the directrix, this indicates that the parabola opens to the right.

Step 2: Form the Equation

The general equation of a parabola that opens right or left is \((y-k)^2 = 4p(x-h)\), where \(p\) is the distance from the vertex to the focus, and \((h, k)\) are the coordinates of the vertex.

Step 3: Insert the Specific Information

In our exercise, the vertex \((h,k)\) is \((0,0)\), and the distance \(p\) is the distance from the vertex to the focus, which is \(\frac{1}{2}\). So, inserting this into the equation gives us \((y-0)^2 = 4*\frac{1}{2}(x-0)\).

Key Concepts

Understanding the Focus and Directrix

When dealing with parabolas, two critical elements are the focus and the directrix. The focus is a point inside the parabola, and the directrix is a line outside it. Each point on the parabola is equidistant from the focus and the directrix. This relationship ensures the symmetrical shape of the parabola.

It's important to remember:

• The focus is not on the parabola itself; it's a guiding point to ensure accurate formation.
• The directrix is a line that helps set the orientation and position of the parabola.
• The parabola will always "open" towards the focus and away from the directrix.

The vertex of the parabola is the midpoint between the focus and the directrix. In our example, the focus is \(\left(\frac{1}{2}, 0\right)\), and the directrix is the line \(x = -\frac{1}{2}\). The vertex lies at \((0,0)\), halfway between these two.

Exploring the Vertex Form Equation

The vertex form equation is a useful way of understanding and graphing parabolas. It is given by:\[(y-k)^2 = 4p(x-h)\]
for parabolas opening right or left, and:\[(x-h)^2 = 4p(y-k)\]
for parabolas opening up or down.

In these formulas:

• \((h, k)\) represents the vertex of the parabola.
• \(p\) is the distance from the vertex to the focus (or the directrix, since they are equidistant).
• "Opening right or left" or "up and down" indicates the direction in which the parabola extends.

In our example, since the parabola opens to the right, we use:\[(y-0)^2 = 4\left(\frac{1}{2}\right)(x-0)\]

This simplifies to:\[y^2 = 2x\]

Steps to Graphing Parabolas

Graphing a parabola involves plotting key points determined by its features, such as focus, vertex, and directrix. Here's how to graph the parabola step by step:

• Identify and plot the vertex on the Cartesian plane. In our example, it's \((0,0)\).
• Mark the focus, \((\frac{1}{2}, 0)\), to the right of the vertex.
• Plot the directrix, which is a vertical line at \(x = -\frac{1}{2}\).
• Draw the parabola ensuring every point is equidistant to the line and focus.
• Double-check the orientation; it opens towards the focus. Since our vertex is at the origin, the parabola will spread symmetrically around it.

Visualizing these components helps in understanding the parabolic structure and mastering its graphing on any coordinate grid.