Problem 1 - 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 \((2,-4)\) $$\frac{x^{2}}{4}+\frac{y^{2}}{16}=1$$

Step-by-Step Solution

Answer

The equation matches with description G.

Step 1: Identify the Type of Conic Section

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{16}=1\). This equation is in the standard form of an ellipse, which is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2\) and \(b^2\) are the denominators. Since both terms are added, it is an ellipse.

Step 2: Determine the Center of the Ellipse

For the standard form of an ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the center is \((h, k)\). Here, there is no \(h\) or \(k\), so the center is \((0, 0)\).

Step 3: Decide the Orientation of the Major Axis

In the ellipse equation, if \(b^2 > a^2\), the major axis is vertical; if \(a^2 > b^2\), it is horizontal. Here, \(a^2 = 4\) and \(b^2 = 16\), so \(b^2 > a^2\), indicating a vertical major axis.

Step 4: Find the Foci of the Ellipse

The foci are given by \((0, \pm c)\), where \(c = \sqrt{b^2-a^2}\). Here, \(c = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}\). Thus, the foci are \((0, \pm 2\sqrt{3})\).

Step 5: Match the Characteristics with the Description

Based on the determined attributes: an ellipse with a vertical major axis, centered at \((0, 0)\) with foci \((0, \pm 2\sqrt{3})\), the equation matches with description G: Ellipse; foci \((0, \pm 2\sqrt{3})\).

Key Concepts

Ellipses

An ellipse is a type of conic section, similar in shape to an elongated circle. It can be formed by slicing a cone with a slanting plane that doesn’t parallel the base or cut through the base of the cone. The resulting shape is symmetrical across both its axes. Ellipses are significant in mathematics and physics because they describe the orbits of planets and satellites. The general form of an equation for an ellipse is given by:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]- Here, - \( (h, k) \) is the center of the ellipse, - \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively.The sum of the distances from any point on the ellipse to the two foci is constant. This unique property helps define the exact shape and orientation of the ellipse. The semi-major axis is the longest diameter of the ellipse, and its orientation (either horizontal or vertical) determines the major axis of the ellipse.

Equation Matching

Equation matching is a mathematical skill that involves comparing an equation to its standard form to determine the type and properties of the conic section represented by the equation. Consider the pair of mathematical expressions:- The standard form of an ellipse is: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]- Here, an equation like \[ \frac{x^2}{4} + \frac{y^2}{16} = 1 \] is an example of an ellipse in its simplified form.The essential steps in equation matching involve:- **Identifying the type** of conic section by comparing with standard forms.- **Calculating the properties** such as center, axes, and orientation.Ultimately, the process allows us to draw meaningful interpretations. In this case, determining whether the major axis is vertical or horizontal by comparing the denominators \( a^2 \) and \( b^2 \). For example, if \( b^2 > a^2 \), then the ellipse has a vertical major axis.

Foci of an Ellipse

The foci (plural of 'focus') of an ellipse are two fixed points on the interior of an ellipse. These points are crucial because they are used to construct and define the ellipse. Any point on the ellipse maintains the property that the sum of the distances from this point to each focus is constant.The formula to find the distance of the foci from the center is:\[ c = \sqrt{b^2 - a^2} \]- In an ellipse where the equation is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \],- If \( b^2 > a^2 \), the major axis is vertical, and the foci are located at \( (h, k \pm c) \).For example, in the ellipse equation:\[ \frac{x^2}{4} + \frac{y^2}{16} = 1 \], we calculate:- \( c = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3} \).- Hence, the foci are at \( (0, \pm 2\sqrt{3}) \).Understanding the position and calculation of the foci helps in grasping how ellipses change and relate to other conic sections such as circles (where the foci coincide).