Problem 13 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ (x-1)^{2}-9(y-4)^{2}=36 $$

Step-by-Step Solution

Answer

The graph is a hyperbola centered at (1, 4), with vertices (7, 4) and (-5, 4).

Step 1: Identify the Equation Type

The given equation \((x-1)^2 - 9(y-4)^2 = 36\) resembles the standard form for a hyperbola, which is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). The presence of a subtraction sign between the squared terms indicates it's a hyperbola.

Step 2: Rewrite in Standard Form

To rewrite the equation in standard form, divide every term by 36 to set the equation equal to 1. \[\frac{(x-1)^2}{36} - \frac{9(y-4)^2}{36} = 1\] Simplify the equation to \[\frac{(x-1)^2}{36} - \frac{(y-4)^2}{4} = 1\]. This matches the standard form of a hyperbola.

Step 3: Determine the Hyperbola's Properties

In the equation \(\frac{(x-1)^2}{36} - \frac{(y-4)^2}{4} = 1\), the center of the hyperbola is at \((h, k) = (1, 4)\). The distance \(a^2 = 36\) gives \(a = 6\), and \(b^2 = 4\) gives \(b = 2\).

Step 4: Graph the Hyperbola

To graph the hyperbola, plot the center at \((1, 4)\). Use the values of \(a\) and \(b\) to define the vertices and asymptotes. The vertices are at \((1 + 6, 4)\) and \((1 - 6, 4)\), which are \((7, 4)\) and \((-5, 4)\). The asymptotes pass through the center with slopes \(\pm \frac{b}{a} = \pm \frac{2}{6} = \pm \frac{1}{3}\). Draw these features on a coordinate plane to represent the hyperbola, with branches opening left and right.

Key Concepts

Standard Form

The standard form of a conic section equation helps identify the type of curve it represents — whether it is a parabola, circle, ellipse, or hyperbola. For hyperbolas, the standard form is \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\],where

• \((h, k)\) is the center of the hyperbola,
• \(a\) is the distance from the center to each vertex on the transverse axis,
• \(b\) is the distance from the center to the asymptotes.

By rearranging a given hyperbola equation to match this form, it becomes easier to analyze its properties and characteristics. Remember to divide every term so that the equation equals 1. This is a key step in converting the equation to the quintessential form expected for detailed analysis.

Asymptotes

Asymptotes of a hyperbola are lines that the hyperbola approaches but never actually meets. They are a crucial part of graphing because they provide the boundary lines that help shape the hyperbola's branches.
For a hyperbola given in the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the asymptotes can be written as:

• \(y - k = \pm \frac{b}{a}(x - h)\)

Here, \(\pm \frac{b}{a}\) represents the slope of the asymptotes, indicating how steeply or gently they lean away from the center point \((h, k)\).
In our case, with \(a = 6\) and \(b = 2\), the slopes are \(\pm \frac{1}{3}\), showing that the asymptotes lean moderately as they extend from the hyperbola's center.

Vertices

The vertices of a hyperbola are the points at which the hyperbola intersects its transverse axis. These points signify the closest distance between the branches of a hyperbola.
Given the standard form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the vertices can be found by:

• Adding and subtracting \(a\) from the \(x\)-coordinate of the center if it's a horizontal hyperbola,
• Adding and subtracting \(a\) from the \(y\)-coordinate of the center for a vertical hyperbola.

In the provided equation, \((x-1)^2/36 - (y-4)^2/4 = 1\), the vertices are located at

• \((h + a, k)\)
• \((h - a, k)\)

which calculate to

• \((7, 4)\)
• \((-5, 4)\)

These points mark the start of the hyperbola's outward path on the graph.

Center of Hyperbola

The center of a hyperbola is a point equidistant from its vertices and the focal points. It serves as the reference point for constructing the rest of the hyperbola's features.
In the standard form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the center is simply given by the coordinates \((h, k)\). Understanding this location is fundamental, as both the directions of the asymptotes and the placement of the vertices are relative to this center point.
For the equation provided, \(\frac{(x-1)^2}{36} - \frac{(y-4)^2}{4} = 1\), the center is

• \((1, 4)\)

This is the baseline position from which all other calculations derive, making it vital for accurately drawing and understanding the hyperbola.