Problem 17 - Contemporary Precalculus

Question

In Exercises \(15-24,\) use the techniques of Examples 4 and 5 to graph the equation in a suitable square viewing window. $$3 x^{2}+2 y^{2}=48$$

Step-by-Step Solution

Answer

Answer: The ellipse has a center at the origin (0, 0), a semi-major axis of 4 units, and a semi-minor axis of \(2\sqrt{6}\) units.

Step 1: Rewrite the equation in standard form

Divide both sides of the equation by \(48\) to get the standard form of the ellipse equation: $$\frac{3x^2}{48} + \frac{2y^2}{48} = 1$$ Simplify the fractions: $$\frac{x^2}{16} + \frac{y^2}{24} = 1$$

Step 2: Determine the center, semi-major axis, and semi-minor axis

From the standard form of an ellipse, \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we can find the center \((h, k)\), the semi-major axis \(a\), and the semi-minor axis \(b\). In our problem, we have: - \(h = 0\) and \(k = 0\). So the center of the ellipse is at the origin, (0, 0). - \(a^2 = 16\). Taking the square root, we get \(a = 4\). This is the semi-major axis along the x-axis. - \(b^2 = 24\). Taking the square root, we get \(b = 2\sqrt{6}\). This is the semi-minor axis along the y-axis.

Step 3: Sketch the ellipse

Now that we have all the information about the ellipse, we can proceed to sketch it: 1. Plot the center of the ellipse at the origin (0, 0). 2. From the center, move 4 units to the right and left along the x-axis. These points are the end vertices of the ellipse. 3. From the center, move \(2\sqrt{6}\) units up and down along the y-axis. These points are the side vertices of the ellipse. 4. Sketch a symmetric curve that passes through all four vertices, creating the ellipse. That's it! The equation \(3x^2 + 2y^2 = 48\) represents an ellipse with the center at the origin, a semi-major axis of 4 units in the x direction, and a semi-minor axis of \(2\sqrt{6}\) units in the y direction.

Key Concepts

Standard Form of an Ellipse

Understanding the standard form of an ellipse is crucial in graphing it accurately. The standard form equation of an ellipse is given by \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] where \((h, k)\) is the center of the ellipse, and \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes respectively. If \(a > b\), the ellipse is stretched further along the x-axis, and if \(b > a\), it extends more along the y-axis.

The given equation, \(3x^2 + 2y^2 = 48\), is not immediately in this standard form. The first step to rewriting it is to divide both sides by the constant term on the right side to normalize the equation. This yields \[\frac{x^2}{16} + \frac{y^2}{24} = 1\] It is now clear that \(h = 0\) and \(k = 0\), indicating that the center of the ellipse is at the origin (0, 0). This puts the equation comfortably in standard form and guides us in determining the lengths of the axes and the subsequent graphing process.

Semi-Major Axis

In an ellipse, the semi-major axis is the longest radius that extends from the center to the perimeter of the ellipse. It's half the length of the major axis, the longest diameter of the ellipse. In the context of the standard form equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] the semi-major axis is represented by \(a\). For our equation \(\frac{x^2}{16} + \frac{y^2}{24} = 1\), we can see that the square of the semi-major axis, \(a^2\), is 16.

To find the semi-major axis, we take the square root and deduce that \(a = 4\). This semi-major axis of 4 units is aligned with the x-axis when graphing the ellipse. By marking points \(4\) units to the left and right of the center, we identify the ends of the major axis on the graph.

Semi-Minor Axis

Conversely, the semi-minor axis is the shortest radius of an ellipse, stretching from the center to the closest edge. It's half the length of the minor axis—which is the shortest diameter. In the standard form equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] \(b\) stands for the length of the semi-minor axis.

For the given ellipse, \(\frac{x^2}{16} + \frac{y^2}{24} = 1\), we find that \(b^2 = 24\). After extracting the square root, we get \(b = 2\sqrt{6}\). This semi-minor axis extends \(2\sqrt{6}\) units above and below the center along the y-axis. By graphing points at these distances from the center, we define the ends of the minor axis. These careful steps in determining the semi-major and semi-minor axes are pivotal for accurately sketching the ellipse's shape.