Problem 75 - A Graphical Approach to College Algebra

Question

Find an equation for each hyperbola. Center \((1,-2)\); focus \((4,-2)\); vertex \((3,-2)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{(x-1)^2}{4} - \frac{(y+2)^2}{5} = 1. \)

Step 1: Determine the Orientation

Given that the center is \((1, -2)\), the vertex is \((3, -2)\), and the focus is \((4, -2)\), we can see that the x-coordinates of these points change while the y-coordinates remain constant. This means that the hyperbola opens horizontally.

Step 2: Find the Transverse Axis Length

In a horizontally opening hyperbola, the transverse axis is along the x-axis. The center \((h, k)\) is \((1, -2)\), and the vertex \((3, -2)\) indicates that \(a\), the distance from the center to the vertex, is \(|3 - 1| = 2\). Thus, \(a = 2\).

Step 3: Calculate the Distance to the Focus

The distance from the center \((1, -2)\) to the focus \((4, -2)\) is \(c = |4 - 1| = 3\). This tells us that \(c = 3\).

Step 4: Determine "b" Using the Relationship

For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is \(c^2 = a^2 + b^2\). Substituting the known values, we have \(3^2 = 2^2 + b^2\). Thus, \(9 = 4 + b^2\) leading to \(b^2 = 5\).

Step 5: Write the Equation of the Hyperbola

The standard form for a horizontally oriented hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Using \((h, k) = (1, -2)\), \(a^2 = 4\) and \(b^2 = 5\), the equation becomes: \[ \frac{(x-1)^2}{4} - \frac{(y+2)^2}{5} = 1. \]

Key Concepts

Understanding the Transverse Axis

The transverse axis is a vital concept in understanding hyperbolas. It's similar to the major axis in an ellipse and indicates the direction along which the hyperbola opens. In this exercise, the hyperbola is defined by the center at \((1, -2)\), vertex at \((3, -2)\), and focus at \((4, -2)\).
The constant y-coordinates and the changing x-coordinates suggest that the transverse axis lies along the x-axis.
For this hyperbola, the transverse axis length is equal to \(2a\), where \(a\) is the distance from the center to a vertex. In this case, we calculate

• \(a = |3 - 1| = 2\)

making the total length of the transverse axis \(2a = 4\).
Understanding this axis helps us know the overall shape and open direction of the hyperbola.

Calculating the Distance to the Focus

For hyperbolas, the focus is a point that is essential for describing the curve's eccentricity. The distance to the focus, labeled as \(c\), is measured from the center of the hyperbola. In the given problem, the center is at \((1, -2)\) and the focus at \((4, -2)\).
This means the focus lies along the transverse axis (the horizontal direction in this case), and the distance can be easily calculated using the x-coordinates:

• \(c = |4 - 1| = 3\)

The value of \(c\) plays a critical role in forming the hyperbola's standard equation and determining how wide or narrow the hyperbola will be.

Formulating the Standard Form Equation

The standard form equation of a hyperbola depends on its orientation, center, and lengths associated with \(a\) and \(b\). For a horizontally opening hyperbola, the standard equation is:
\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]
where \((h, k)\) represents the center.
Using \((h, k) = (1, -2)\), \(a^2 = 4\) as calculated from the distance to the vertex, and \(c = 3\), we can apply the formula \(c^2 = a^2 + b^2\) to find \(b^2\):

• \(9 = 4 + b^2\)
• Thus, \(b^2 = 5\)

Inserting all these values gives us the equation:
\[\frac{(x-1)^2}{4} - \frac{(y+2)^2}{5} = 1\]
This equation describes the hyperbola completely in its standard form.