Problem 51 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$x=\frac{1}{2}(y-1)^{2}-2$$

Step-by-Step Solution

Answer

The parabola is \((y-1)^2 = 2(x+2)\) with vertex \((-2,1)\), and it opens to the right.

Step 1: Understand the Equation Type

The given equation is \( x = \frac{1}{2}(y-1)^2 - 2 \). It is in the form of a parabola because it has a squared term and one of the variables, \( y \), is squared, indicating a horizontal parabola.

Step 2: Convert to Standard Form of a Parabola

The standard form of a horizontal parabola is \((y-k)^2 = 4p(x-h)\). To write \( x = \frac{1}{2}(y-1)^2 - 2 \) in this form, first isolate the squared term: \( x + 2 = \frac{1}{2}(y-1)^2 \). Multiply both sides by 2 to simplify: \((y-1)^2 = 2(x + 2) \). Now it's in the form \((y-k)^2 = 4p(x-h)\), where \(h = -2\), \(k = 1\), and \(4p = 2\), leading to \(p = \frac{1}{2}\).

Step 3: Identify Key Elements

The vertex of the parabola is at \((h, k) = (-2, 1)\). The focus is \(\left(h+p, k\right) = \left(-2 + \frac{1}{2}, 1\right) = \left(-\frac{3}{2}, 1\right)\). The directrix is a vertical line at \(x = h - p = -\frac{5}{2}\).

Step 4: Graph the Parabola

Plot the vertex at \((-2, 1)\). Then draw the parabola, opening to the right because the \(p\) value is positive. Ensure the focus \((-\frac{3}{2}, 1)\) is inside the curve and the directrix \(x = -\frac{5}{2}\) is outside, showing the parabola's symmetric nature about the line y = 1.

Key Concepts

Horizontal Parabola

A parabola is a two-dimensional, mirror-symmetrical curve that is shaped like an arch. A unique feature of a horizontal parabola, as opposed to a vertical one, is that it opens sideways rather than up or down. This particular orientation occurs when the squared term is associated with the vertical axis, which means the equation

• is typically expressed in the form \[(y-k)^2 = 4p(x-h)\]

where

• \(h\) and \(k\) indicate the coordinates of the vertex, and
• \(p\) determines the distance from the vertex to the focus and directrix.

In the given equation, \((y-1)^2 = 2(x + 2)\), we can see the format matches, highlighting it's a horizontal parabola. Understanding this ensures that when graphing, we anticipate the curve to open to the right for positive \(p\) or to the left for negative \(p\).

Vertex

The vertex serves as a central point of a parabola from which the curve emanates. It is where the parabola changes direction and is often considered the minimum or maximum point of the function, depending on the parabola's orientation.
For our horizontal parabola, the vertex can be located directly from the standard form
\((y-k)^2 = 4p(x-h)\), where the vertex is found at the coordinates
\((h, k)\).In the equation \((y-1)^2 = 2(x + 2)\), rewriting it reveals that

• \(h = -2\), and
• \(k = 1\).

Thus, the vertex is at

• \((-2, 1)\)

showing where the axes of symmetry intersect the curve. When plotting, this point gives a crucial reference to shape the parabola accurately, with the entire curve symmetrical about the horizontal line \(y = k\).

Focus

The focus of a parabola is a pivotal reference point that helps define the curve's shape. It is located inside the parabola, along with its axis of symmetry. This point, in conjunction with the directrix, establishes the locus of all points equidistant from both the focus and directrix, forming the parabola.The formula for finding the focus, given the vertex

• \((h, k)\)

is by moving along the axis of the parabola by a distance

• \(p\)

units from the vertex. In our equation with

\(p = \frac{1}{2}\)

, the focus

• lies at \((h+p, k) = \left(-2 + \frac{1}{2}, 1\right)\).

This results in a focus

• located at \(\left(-\frac{3}{2}, 1\right)\).

When graphing, ensure this point is correctly placed inside the parabola, as it dictates the direction and nature of the opening curve.

Directrix

The directrix is as vital as the focus, serving as a guide for the parabola. It is a line that helps ensure all points on the parabola are equidistant from itself and the focus. For a horizontal parabola, the directrix is a vertical line.To determine the directrix of our given equation, use:

• \(x = h - p\)

Given

• \(p = \frac{1}{2}\)

, the line is at:

• \(x = -2 - \frac{1}{2} = -\frac{5}{2}\).

This vertical line should be drawn outside the parabola, opposite the direction of the focus, thus confirming the symmetrical nature of the parabola about the axis \(y = 1\). When you plot this, always maintain it as a line the parabola never touches, providing a boundary and shape reference for plotting each parabola point.