Problem 58 - College Algebra

Question

Express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(\frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1\)

Step-by-Step Solution

Answer

\(y = -3 \pm \sqrt{\frac{25}{16}((x-2)^{2} - 16)}\): two functions for the hyperbola.

Step 1: Identify the form of the hyperbola equation

The given equation \(\frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1\) is in the standard form of a hyperbola centered at \((h, k)\), where \(a^2=16\) and \(b^2=25\). Identify these values to transform \(y\) as a function of \(x\).

Step 2: Solve for \(y\) in terms of \(x\)

To express \(y\) as a function of \(x\), isolate the term involving \(y\):\[\frac{(y+3)^{2}}{25} = \frac{(x-2)^{2}}{16} - 1\].Next, simplify the right-hand side: \[\frac{(y+3)^{2}}{25} = \frac{(x-2)^{2} - 16}{16}\].

Step 3: Eliminate the fraction

Multiply both sides by 25 to eliminate the fraction:\[(y+3)^{2} = \frac{25}{16}((x-2)^{2} - 16)\].This step prepares the equation for extracting the square root to solve for \(y\).

Step 4: Isolate \((y+3)\)

Take the square root of both sides to solve for \(y+3\):\[y+3 = \pm \sqrt{\frac{25}{16}((x-2)^{2} - 16)}\].This gives two potential functions based on the sign of the square root.

Step 5: Express \(y\)

Finally, express \(y\) as a function of \(x\):\[y = -3 \pm \sqrt{\frac{25}{16}((x-2)^{2} - 16)}\].This represents the two branches of the hyperbola.

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola equation is important because it reveals the hyperbola's shape and orientation. For a hyperbola centered at \((h, k)\), the equation is given as:

• \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
• or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)

The difference in these equations signifies the hyperbola's orientation—horizontal or vertical. If \(x\) is positive, the hyperbola opens left and right, whereas if \(y\) is positive, it opens up and down.

Knowing \(a^2\) and \(b^2\), you can determine the lengths of the transverse and conjugate axes. In our exercise, the given equation is \(\frac{(x-2)^2}{16} - \frac{(y+3)^2}{25} = 1\), indicating a horizontally oriented hyperbola with center at \((2, -3)\). Here, \(a^2=16\) and \(b^2=25\), which helps in graphing and interpreting the hyperbola.

Graphing Calculator

Using a graphing calculator to visualize hyperbolas can greatly aid understanding. Start by inputting the equation in standard form. In our example, we rewrote the equation as two functions: \\(y = -3 \pm \sqrt{\frac{25}{16}((x-2)^2 - 16)}\).

Here's how to use a graphing calculator:

• Enter the two separate functions: one with the '+', another with the '-'. This displays both branches of the hyperbola.
• Ensure your calculator is set to use an appropriate viewing window, likely centering around the hyperbola's center \((2, -3)\).
• Adjust the view to encompass the curves visually; experiments with the scale may be necessary to capture both branches clearly.
• Utilize the calculator's zoom and tracing capabilities to better understand the hyperbola's structure and intersections.

Graphing provides a tangible image of the hyperbola, corroborating its mathematical description.

Equation of a Hyperbola

Understanding the equation is key to exploring different aspects of a hyperbola. As mentioned, the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or vice versa, dictates the hyperbola’s properties.

The position \((h, k)\) is the center, while \(a\) and \(b\) define the distance from the center to the vertices and the distance from the center to the co-vertices, respectively. The branches of a hyperbola never meet, extending infinitely,

• opening around asymptotes defined by \( y = k \pm \frac{b}{a}(x-h) \) or \(x = h \pm \frac{b}{a}(y-k)\).

In our exercise, transforming the equation to \\(y = -3 \pm \sqrt{\frac{25}{16}((x-2)^2 - 16)}\) unravels these properties, providing concrete transformations essential for both algebraic manipulation and graphing. Examining the hyperbolic equation offers insight into geometric assignments, like foci and eccentricity.

Functions of a Hyperbola

Hyperbolas can be split into two separate functions, each representing a branch of the hyperbola. In our example, rewriting the equation gives

• \(y = -3 + \sqrt{\frac{25}{16}((x-2)^2 - 16)}\)
• \(y = -3 - \sqrt{\frac{25}{16}((x-2)^2 - 16)}\)

Each equation represents one of the two branches of the hyperbola, visually apparent when graphing.

These functions describe how \(y\) depends on \(x\) for either branch, accounting for the different signs in the square root term. This approach emphasizes the hyperbola's symmetry and helps clarify its behavior as \(x\) deviates from the center.

Breaking the hyperbola down like this reveals its underlying structure and offers a clearer perspective when using graphing calculators, solving algebraically, or examining intersections.