Problem 14 - 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. \(x=\frac{1}{8} y^{2}\)

Step-by-Step Solution

Answer

The vertex is \((0, 0)\), focus is \((2, 0)\), and directrix is \(x = -2\).

Step 1: Identify the Equation Type

The given equation is in the form \( x = \frac{1}{4c} y^2 \), which is the standard form of a parabola that opens to the right or left. In this case, since \( \frac{1}{4c} = \frac{1}{8} \), the parabola opens to the right.

Step 2: Rewrite in Standard Form

The standard form of a parabola with a horizontal axis that opens to the right is given by \( (y-k)^2 = 4p(x-h) \). Comparing it to \( x = \frac{1}{8} y^2 \), we can see \( h = 0 \) and \( k = 0 \), so the equation \( y^2 = 8x \) fits the form \( y^2 = 4px \), with \( 4p = 8 \), thus \( p = 2 \).

Step 3: Determine the Vertex

For the parabola \( y^2 = 4p(x-h) \), the vertex \( V \) is at \( (h, k) \). From our equation, \( h = 0 \) and \( k = 0 \), so the vertex of the parabola is at \( (0, 0) \).

Step 4: Determine the Focus

The focus \( F \) of the parabola is located \( p \) units from the vertex along the axis of symmetry. Since \( p = 2 \), and the parabola opens to the right, the focus is at \( (h+p, k) = (2, 0) \).

Step 5: Determine the Directrix

The directrix of the parabola is a vertical line located \( p \) units from the vertex in the opposite direction of the focus. Since \( p = 2 \), and the parabola opens to the right, the directrix is \( x = -p \), thus \( x = -2 \).

Key Concepts

Vertex

In the context of a parabola, the vertex is a crucial point. It acts as a kind of 'anchor' that defines the curve's position on a graph. A parabola can open upwards, downwards, or sideways, but it always revolves around its vertex.

The vertex represents the midpoint along the axis of symmetry, giving a parabola its balanced shape. You can think of it as the 'turning point' where the curve changes direction. It is defined by the coordinates \(h, k\). In the case of the equation \(x = \frac{1}{8}y^2\), the vertex is at \(0, 0\).

• This happens because the terms \(h = 0\) and \(k = 0\) in the standard form equation \(y^2 = 4p(x-h)\) place our vertex at the origin.
• Knowing the vertex is essential as it often simplifies other calculations, like finding the focus or plotting the parabola.

Focus

The focus of a parabola is another key component that shapes its form. Unlike the vertex, the focus is not visible on the parabola graph itself but plays a vital role in its geometry.

The focus sits inside the parabolic curve and is \(p\) units away from the vertex, influencing the direction in which the parabola opens. It helps determine the breadth and orientation of the curve.

• For the parabola given in the equation \(y^2 = 4p(x-h)\), \(p\) is found to be 2 from solving \(4p = 8\).
• Since our parabola opens to the right from its vertex \(0, 0\), the focus will be at the point \(2, 0\).
• This location means the parabola is 'stretched' towards the focus, defining the curve's concave nature.

Directrix

The directrix is a line associated with the parabola that contrasts with the focus. It is not a part of the parabola itself but crucial to its definition and construction. The directrix lies perpendicular to the axis of symmetry and is located \(p\) units from the vertex, opposite the focus. It helps ensure that every point on the parabola is equidistant from the focus and the directrix.

• Working with the equation \(y^2 = 4p(x-h)\), where \(p = 2\), we find that the directrix is the line \(x = -2\).
• This means, just as the parabola stretches out towards its focus at \(2, 0\), its opposite boundary is set by this vertical line, the directrix at \(x = -2\).
• Understanding the directrix's role provides more insight into how the parabola is structured around its central elements, the vertex and the focus.