Problem 8 - A Graphical Approach to Precalculus with Limits

Question

Match each equation in Column I with the appropriate description in Column II. Do not use a calculator. \(\mathbf{II}\) 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

The equation matches description A: Hyperbola; center \((2, 4)\).

Step 1: Identify the Conic Section

Let's analyze the given equation: \( \frac{(x-2)^2}{9} - \frac{(y-4)^2}{25} = 1 \). This is in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), which represents a hyperbola with a horizontal transverse axis.

Step 2: Determine the Center of the Hyperbola

In the standard form of a hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center is \((h, k)\). From the equation \( \frac{(x-2)^2}{9} - \frac{(y-4)^2}{25} = 1 \), we identify \(h = 2\) and \(k = 4\). Thus, the center is \((2, 4)\).

Step 3: Match the Correct Description

Locate the description in Column II that matches a hyperbola with the center \((2, 4)\). Looking at the descriptions provided, description A: "Hyperbola; center \((2, 4)\)" fits the information derived from the equation.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that you can easily spot using its unique equation form. The general equation for a hyperbola is given by \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] if it opens along the x-axis, or \[ \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \] if it opens along the y-axis. The equation from the problem \[ \frac{(x-2)^2}{9} - \frac{(y-4)^2}{25} = 1 \] shows that the hyperbola opens along the x-axis because the term with \((x-2)^2\) is positive.Key characteristics of a hyperbola include:

• Center: The midpoint between the two foci, given by \((h, k)\).
• Axes: A hyperbola can have either a horizontal or vertical transverse axis, depending on the equation.
• Foci: Points located along the transverse axis, found using the formula \(c = \sqrt{a^2 + b^2}\).

By recognizing the hyperbola's equation structure, you can quickly determine important features like its orientation and center.

Ellipse

An ellipse is another fascinating conic section characterized by its "stretched circle" shape. The general form of an ellipse's equation is \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where \(a\) and \(b\) determine the ellipse's width and height, respectively. If \(a > b\), the ellipse is wider; if \(b > a\), it's taller.Here's what you should remember about ellipses:

• Center: Like a hyperbola, the center is at \((h, k)\).
• Axes: The major axis is the longest diameter; the minor axis is the shortest.
• Foci: Located on the major axis, found using \(c = \sqrt{a^2 - b^2}\).

An ellipse with a horizontal major axis stretches along the x-axis, while a vertical major axis elongates along the y-axis. Identifying these features is key when matching equations to descriptions!

Center

The concept of the center is crucial when studying conic sections like hyperbolas and ellipses. In mathematics, the center of a conic section is often denoted as \((h, k)\), representing the point around which the shape is symmetrically located.To find the center:

• In a hyperbola or ellipse in standard form, the center is directly extracted from the equation's terms \((x-h)^2\) and \((y-k)^2\).
• For instance, in the hyperbola equation \(\frac{(x-2)^2}{9} - \frac{(y-4)^2}{25} = 1\), the center is \((2, 4)\).
• Understanding where a conic section is centered helps in sketching it correctly and linking it to the appropriate description.

Remember, knowing the center gives you a clear starting point for exploring the shape and placement of conic sections on the coordinate plane.