Problem 58 - College Algebra

Question

For the following exercises, 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

Two functions are \( y_1 = -3 + \sqrt{\frac{25}{16}(x-2)^2 - 25} \) and \( y_2 = -3 - \sqrt{\frac{25}{16}(x-2)^2 - 25} \).

Step 1: Identify the Standard Form of the Hyperbola

The given equation is \( \frac{(x-2)^{2}}{16} - \frac{(y+3)^{2}}{25} = 1 \). This is in the standard form of a hyperbola with horizontal transverse axis:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]where \( h = 2 \), \( k = -3 \), \( a^2 = 16 \), and \( b^2 = 25 \). "Step 2: Solve for y in terms of x: Extract and RearrangeCancel small elements, we will correct stile cancel."

Step 2: Solve for y in terms of x: Isolate (y + 3)^2

Rearrange the equation to isolate \( (y + 3)^2 \):\[ \frac{(x-2)^2}{16} - 1 = \frac{(y+3)^2}{25} \]Simplify the left side:\[ \frac{(x-2)^2}{16} - \frac{16}{16} = \frac{(y+3)^2}{25} \]\[ \frac{(x-2)^2 - 16}{16} = \frac{(y+3)^2}{25} \]

Step 3: Cross Multiply to Clear Fractions

Cross-multiply to clear fractions:\[ 25((x-2)^2 - 16) = 16(y+3)^2 \]Expand on the left:\[ 25(x-2)^2 - 400 = 16(y+3)^2 \]

Step 4: Solve for y: Rearrange and Simplify

Rearrange to solve for \( y \):\[ 16(y+3)^2 = 25(x-2)^2 - 400 \]Divide by 16:\[ (y+3)^2 = \frac{25}{16}(x-2)^2 - 25 \]Take the square root on both sides:\[ y + 3 = \pm \sqrt{\frac{25}{16}(x-2)^2 - 25} \]Subtract 3:\[ y = -3 \pm \sqrt{\frac{25}{16}(x-2)^2 - 25} \]

Step 5: Express as Two Functions

This expression for \( y \) gives us two functions, representing the upper and lower halves of the hyperbola:\[ y_1(x) = -3 + \sqrt{\frac{25}{16}(x-2)^2 - 25} \]\[ y_2(x) = -3 - \sqrt{\frac{25}{16}(x-2)^2 - 25} \]

Key Concepts

Standard Form of a Hyperbola

To express the equation of a hyperbola correctly, we often use its standard form. The standard form helps us identify key characteristics and details about the hyperbola. In this context, we are given \( \frac{(x-2)^{2}}{16} - \frac{(y+3)^{2}}{25} = 1 \). This equation is already in the standard form of a hyperbola, which is: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
Here, the hyperbola is horizontal because the term with \(x\) is positive and comes first. For our given equation, the values are:

• \(h = 2\)
• \(k = -3\)
• \(a^2 = 16\)
• \(b^2 = 25\)

Understanding these components, \((h, k)\), gives us the center of the hyperbola at \((2, -3)\). The values \(a^2\) and \(b^2\) define the distance to the vertices and co-vertices from the center along the axes.

Solving Equations for y

Solving for \(y\) in terms of \(x\) is a key part of understanding the behavior of a hyperbola. Our goal is to express \(y\) as two separate functions of \(x\) since a hyperbola has two branches.
Starting with the equation \(\frac{(x-2)^{2}}{16} - \frac{(y+3)^{2}}{25} = 1\), we rearrange to isolate \((y+3)^2\):

• Subtract and simplify: \(\frac{(x-2)^2}{16} - 1 = \frac{(y+3)^2}{25}\).
• Simplify further: \(\frac{(x-2)^2 - 16}{16} = \frac{(y+3)^2}{25}\).

This equation tells us that after rearranging and removing fractions via cross-multiplication and simplification, we arrive at:
\( (y+3)^2 = \frac{25}{16}(x-2)^2 - 25 \).
Finally, taking the square root and solving for \(y\), we get two functions:

• \(y_1(x) = -3 + \sqrt{\frac{25}{16}(x-2)^2 - 25}\)
• \(y_2(x) = -3 - \sqrt{\frac{25}{16}(x-2)^2 - 25}\)

These represent the two parts or "branches" of the hyperbola.

Graphing Calculator

Using a graphing calculator is a practical tool to visualize the hyperbola, especially when it involves more complex algebraic manipulations. A graphing calculator can help verify our solutions and make sure that the algebraic expressions we found accurately represent the hyperbola.
Once we have expressed the hyperbola in two functions, \(y_1(x)\) and \(y_2(x)\), follow these steps to graph them:

• Enter \(y_1(x) = -3 + \sqrt{\frac{25}{16}(x-2)^2 - 25}\) into the calculator.
• Enter \(y_2(x) = -3 - \sqrt{\frac{25}{16}(x-2)^2 - 25}\) as a separate function.

Make sure to use a suitable window setting that captures the behavior of the hyperbola. Typically, you may adjust the x-range around the center \(h = 2\) and similar adjustments vertically around \(k = -3\).
This tool allows for real-time changes and can help identify any errors in calculation, enhancing understanding and ensuring the integrity of the graph.