Problem 15 - Intermediate Algebra

Question

Find \(h, k, a,\) and \(b: \frac{(x-5)^{2}}{25}-\frac{(y+11)^{2}}{36}=1\)

Step-by-Step Solution

Answer

The values are \( h = 5 \), \( k = -11 \), \( a = 5 \), \( b = 6 \).

Step 1: Identify the Standard Form

The given equation is \( \frac{(x-5)^{2}}{25}-\frac{(y+11)^{2}}{36}=1 \). This is the standard form of a hyperbola equation, \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \), which opens horizontally.

Step 2: Compare and Extract the Values

By comparing the given equation \( \frac{(x-5)^{2}}{25}-\frac{(y+11)^{2}}{36}=1 \) with the standard form \( \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \), we can observe: \( h = 5 \), \( k = -11 \), \( a^{2} = 25 \), and \( b^{2} = 36 \).

Step 3: Calculate Values of \(a\) and \(b\)

Since \( a^{2} = 25 \), take the square root to find \( a = \sqrt{25} = 5 \). Similarly, \( b^{2} = 36 \), so \( b = \sqrt{36} = 6 \).

Key Concepts

Standard Form of Hyperbola

To understand the concept of hyperbolas, it's important to start with their standard form equation. This form provides a template that makes it easy to identify characteristic values from the equation itself, such as the center, vertices, and more. In the case of hyperbolas, the standard form of the equation looks like this: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Here, the variables \( h \) and \( k \) signify the center of the hyperbola. The values \( a \) and \( b \) are fundamental in determining the dimensions and orientation of the hyperbola.

• \( h \) and \( k \) provide the horizontal and vertical translation from the origin, respectively, forming the center \((h, k)\).
• \( a^2 \) and \( b^2 \) are the denominators, directly affecting the shape and measurement of the axes of the hyperbola.

Notably, in this standard form, the hyperbola opens horizontally if the positive term is associated with the \( x \) variable, as showcased in this problem, or vertically if it's linked to the \( y \) variable. This preliminary knowledge is crucial for comprehending more complex modifications and transformations.

Identifying Values in Equations

Once the standard form of a hyperbola is recognized, the next step involves extracting useful values that describe the hyperbola's features. When presented with an equation like:\[ \frac{(x-5)^2}{25} - \frac{(y+11)^2}{36} = 1 \]We proceed by matching it with the structure of the standard form. The placement and values of each component in the equation help define the respective parameters:

• Here, \( h \) is given as 5, derived from replacing \( x \) within the translation component \((x-h)^2\).
• Similarly, \( k \) is \(-11\), extracted from \((y-k)^2\).
• The value \( a^2 \) is found as 25, from the denominator of the \( x \) term, corresponding to \( a = 5 \).
• The value \( b^2 \) stands at 36, tied to the \( y \) denominator, thus \( b = 6 \).

Applying these values back to the hyperbola's description in terms of its center and axes allows us to visualize and interpret the hyperbola effectively. Identifying and comprehending these extracted values make solving and graphing hyperbolic equations not only approachable but a routine task.

Equation Transformation

Equation transformation is a critical skill to ensure comprehension of how changes in equations affect the graph of a hyperbola. Transformations involve shifting, stretching, or flipping the graph, corresponding to different modifications in the equation parameters. Starting from the standard form:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Knowing how alterations like translating \( h \) and \( k \) values will move the hyperbola’s center horizontally and vertically gives us substantial insight into graphing. Yet, transformations go beyond simple shifts:

• Adjusting values of \( a \) or \( b \) squeezes or stretches the hyperbola along its axes, affecting how ‘open’ or ‘tight’ the curves are.
• Swapping roles of \( a^2 \) and \( b^2 \) changes whether the hyperbola opens upwards (vertically) or sideways (horizontally).

Understanding these transformations allows for flexible manipulation of equations to predict and achieve desired graph shapes and positions. Grasping this level of transformation is invaluable when taking equations from abstract numbers to tangible visual representations.