Problem 7 - A Graphical Approach to College Algebra

Question

Match each equation in Column I with the appropriate description in Column II. Do not use a calculator. A. Hyperbola; center \((2,4)\) B. Ellipse; foci \((\pm 2 \sqrt{3}, 0)\) C. Hyperbola; foci \((0, \pm 2 \sqrt{5})\) D. Hyperbola; center \((-2,4)\) E. Ellipse; center \((-2,4)\) F. Center \((0,0) ;\) horizontal transverse axis G. Ellipse; foci \((0, \pm 2 \sqrt{3})\) H. Vertical major axis; center \((2,-4)\) $$\frac{(x+2)^{2}}{9}-\frac{(y-4)^{2}}{25}=1$$

Step-by-Step Solution

Answer

D. Hyperbola; center \((-2,4)\)

Step 1: Identify the Type of Equation

The given equation is \( \frac{(x+2)^{2}}{9} - \frac{(y-4)^{2}}{25} = 1 \). This is an equation of a hyperbola because it is of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). The negative sign indicates it represents a hyperbola.

Step 2: Determine the Center

The general form of the hyperbola equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) shows that the center is at \((h, k)\). For our equation, \(h = -2\) and \(k = 4\), so the center of the hyperbola is \((-2, 4)\).

Step 3: Match with Column II Options

From the given descriptions, D. "Hyperbola; center \((-2,4)\)" matches our findings. Thus, the given equation corresponds to this description.

Key Concepts

Hyperbola

A hyperbola is a fascinating type of conic section created by the intersection of a plane with a double cone. If you think of slicing through a cone with a knife, the shape you form is what we call a conic section.

Unlike an ellipse, which forms an oval shape, a hyperbola consists of two symmetrical branches. These branches open either horizontally or vertically and are mirrored along an axis. The recognition of a hyperbola comes from its equation, which features subtraction between two squared terms.

In the standard form, a hyperbola's equation looks like this: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] The negative sign between the terms is key. It allows us to identify the equation's nature as a hyperbola, differentiating it from other conic sections like ellipses and circles. Understanding this can help you swiftly recognize hyperbolas in any exercise.

Center of Hyperbola

The center of a hyperbola is a crucial point that defines its position in the coordinate plane. Think of it as the point of balance for the two branches of the hyperbola. In the equation of a hyperbola, \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] , the center is represented by the coordinates \((h, k)\).

Identifying the center is straightforward. Simply take the values directly from the equation: \((h, k)\). In our problem, we had \((h, k) = (-2, 4)\), meaning the center is located at the point \((-2, 4)\) on the graph.

Knowing the center helps in sketching the graph and understanding other properties of the hyperbola, such as its orientation and axis. Always start by pinpointing the center to ground your understanding before moving on to other attributes.

Equation of Hyperbola

The equation of a hyperbola holds the key to understanding its structure and visual representation. In general, the equation can take two forms, each indicating a different orientation:

• Horizontal Hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• Vertical Hyperbola: \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \)

The choice of form depends on whether the hyperbola opens sideways or upwards and downwards. The negative sign always indicates the nature of the conic section as a hyperbola.

In our particular exercise, we dealt with the equation: \[ \frac{(x+2)^2}{9} - \frac{(y-4)^2}{25} = 1 \] This identifies a hyperbola with a horizontal transverse axis because the \((x-h)^2\) term comes first. The coefficients under these squared terms, \(9\) and \(25\), play a vital role in determining the hyperbola's shape and size. They represent \(a^2\) and \(b^2\), respectively, and affect the distance between various parts of the hyperbola. Recognizing these elements in an equation aids in visualizing and sketching the graph accurately.