Problem 12 - College Algebra

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. $$ y=\frac{1}{4} x^{2} $$

Step-by-Step Solution

Answer

Vertex: (0,0), Focus: (0,1), Directrix: y = -1.

Step 1: Identify Current Form and Requirement for Standard Form

The given equation is \( y = \frac{1}{4} x^{2} \). This represents a parabola. Our task is to express it in standard form for a parabola that opens up or down, which is \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex.

Step 2: Convert the Equation to Standard Form

The equation \( y = \frac{1}{4} x^2 \) is already in the standard form of \( y = a(x-h)^2 + k \) by recognizing \( a = \frac{1}{4} \), \( h = 0 \), and \( k = 0 \). Therefore, \( (h, k) = (0, 0) \), making this the vertex.

Step 3: Determine the Vertex

From the standard form \( y = \frac{1}{4} x^2 \), we find that the vertex \( (V) \) is \( (0, 0) \).

Step 4: Determine the Focus

For a parabola \( y = a(x-h)^2 + k \), the focus can be found using the formula \((h, k + \frac{1}{4a})\). Here, \( a = \frac{1}{4} \), making \( \frac{1}{4a} = 1 \). So, the focus is at \( (0, 0 + 1) = (0, 1) \).

Step 5: Determine the Directrix

The directrix of a parabola \( y = a(x-h)^2 + k \) is given by \( y = k - \frac{1}{4a} \). Substituting \( a = \frac{1}{4} \) and \( k = 0 \), we have \( y = 0 - 1 = -1 \). Thus, the directrix is \( y = -1 \).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key feature, as it is the point that depicts the parabola's most extreme value. If the parabola opens upwards, the vertex is the lowest point. Conversely, for a downward-opening parabola, it becomes the highest point. Understanding how to locate the vertex helps us better grasp the parabola's shape and position.
For parabolas in the form \( y = a(x-h)^2 + k \), the vertex can be identified as \((h, k)\). In our specific equation \( y = \frac{1}{4} x^2 \), by comparison with the standard vertex form, it can be established that \( h = 0 \) and \( k = 0 \). This results in the vertex at the origin: \((0, 0)\).

• The vertex indicates where the parabola changes direction from increasing to decreasing or vice versa.
• It helps to determine the axis of symmetry, which runs vertical to the vertex through \( x = h \).
• Since our parabola opens upwards, the vertex is a minimum point.

Focus of a Parabola

The focus of a parabola is an internal point lying on its axis of symmetry, which plays a significant role in how the parabola reflects light and sound. Imagine drawing a line directly to this point from anywhere on the parabola's curve—the focus dictates their direction.
In the standard form \( y = a(x-h)^2 + k \), the focus is calculated using \((h, k + \frac{1}{4a})\). For the equation \( y = \frac{1}{4} x^2 \), we identify that \( a = \frac{1}{4} \), hence \( \frac{1}{4a} = 1 \). This means our focus is located at position \((0, 1)\).

• The focus is always above the vertex for parabolas that open upwards and below for those opening downwards.
• It assists in problems involving reflections and paths of projectiles.
• The position of the focus indicates how "steep" or "flat" the parabola is.

Directrix of a Parabola

The directrix is a horizontal line outside of the parabola that helps define its shape. Every point on the parabola is equidistant from both the focus and this line.
For a standard equation \( y = a(x-h)^2 + k \), the directrix is determined by \( y = k - \frac{1}{4a} \). Given \( a = \frac{1}{4} \) and \( k = 0 \) in our equation \( y = \frac{1}{4} x^2 \), the directrix settles at \( y = -1 \).

• It is always parallel to the axis of symmetry.
• In conjunction with the focus, it perfectly defines a parabola's trajectory through a geometric property called the "locus."
• Understanding the position of the directrix aids in comprehension of the parabola's width and steepness.