Problem 13 - College Algebra with Corequisite Support

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. \(y=-4 x^{2}\)

Step-by-Step Solution

Answer

The vertex is (0,0), the focus is (0,-1/16), and the directrix is y=1/16.

Step 1: Convert to Standard Parabola Form

The given equation is \( y = -4x^2 \). The standard form for a vertical parabola is \( (x-h)^2 = 4p(y-k) \). We want to rewrite \( y = -4x^2 \) to match this form. We start by rearranging it to get \( x^2 = -\frac{1}{4}y \).

Step 2: Identify Parameters from Standard Form

In the rearranged form \( x^2 = -\frac{1}{4}y \), compare it with \( x^2 = 4p(y-k) \). Here, \( 4p = -\frac{1}{4} \), so \( p = -\frac{1}{16} \). The vertex (h, k) is at \((0,0)\) since there are no shifts in \(x\) or \(y\).

Step 3: Determine the Vertex

From the parameters identified, the vertex \( V \) of the parabola is \( (h, k) = (0, 0) \).

Step 4: Determine the Focus

Using \( p = -\frac{1}{16} \) and the vertex \((0,0)\), the focus \( F \) of the parabola is \( (0, k + p) = (0, -\frac{1}{16}) \).

Step 5: Determine the Directrix

The directrix \( d \) of the parabola is a horizontal line at \( y = k - p \). Substituting \( k = 0 \) and \( p = -\frac{1}{16} \), the directrix is \( y = \frac{1}{16} \).

Key Concepts

Standard Form of a Parabola

Understanding the standard form of a parabola is essential for correctly identifying key features like the vertex, focus, and directrix. For parabolas that open either upwards or downwards, the standard form is expressed as: \ \( (x-h)^2 = 4p(y-k) \) \ In this formula: \

• \((h, k)\) represent the coordinates of the vertex, which is the highest or lowest point of the parabola depending on its direction.
• \( p \) is a crucial parameter related to the parabola's width and direction of opening.

To convert a general equation like \( y = -4x^2 \) into this form, we rearrange it to isolate \( x^2 \) on one side. This process aligns the equation with the standard form by adjusting coefficients accordingly.

Vertex of a Parabola

The vertex of a parabola is the point where it turns or changes direction. It is a key feature that gives us critical information about the parabola's position on the graph. In the equation \ \( (x-h)^2 = 4p(y-k) \) \ The vertex is represented by \((h, k)\). When the original equation is \( y = -4x^2 \), the vertex is found at \((0, 0)\). This is because there are no shifts occurring in the \(x\) or \(y\) direction, retaining its central position at the origin. The vertex essentially describes the central anchor point of the parabola on the coordinate plane.

Focus of a Parabola

The focus of a parabola is another vital component that defines its shape and direction. It is one of the points that helps determine the "thickness" of the parabola. In our specific example, we derived the focus from the value of \( p \), where \( p = -\frac{1}{16} \). \ With the vertex already identified as \((0,0)\), the focus \((F)\) is located at \((0, k + p)\), giving us the point \((0, -\frac{1}{16})\). The focus lies along the axis of symmetry of the parabola and is used, along with the directrix, to define the parabola's reflective property.

Directrix of a Parabola

The directrix of a parabola is an imaginary line that helps to define its structure. It is positioned opposite the parabola from the focus, relative to the vertex. Whereas the focus is a point, the directrix is a line that plays a critical role in delineating the parabola's form. \ In our problem, the directrix is calculated using \( y = k - p \). Since \( k \) is \(0\) and \( p \) is \(-\frac{1}{16}\), the equation becomes \( y = \frac{1}{16} \). This line is parallel to the x-axis and assists in maintaining the shape by balancing the influence of the focus on the parabola.