Problem 14 - A Graphical Approach to Precalculus with Limits

Question

Graph each ellipse by hand. Give the domain and range. Give the foci in Exercises \(11-14\) and identify the center in Exercises \(17-22 .\) Do not use a calculator. $$12 x^{2}+8 y^{2}=96$$

Step-by-Step Solution

Answer

Center: (0,0); Foci: (0,±2); Domain: [-2√2, 2√2]; Range: [-2√3, 2√3].

Step 1: Write the Equation of the Ellipse

The standard form of the equation for an ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). To match the given equation \( 12x^2 + 8y^2 = 96 \) to the standard form, first divide the entire equation by 96: \( \frac{12x^2}{96} + \frac{8y^2}{96} = \frac{96}{96} \), which simplifies to \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \).

Step 2: Identify the Center of the Ellipse

Since the equation \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \) is centered at the origin because it has the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where both \( x \) and \( y \) are not translated by any constant, the center of the ellipse is at (0,0).

Step 3: Determine the Axes' Lengths

From the equation \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \), compare the denominators to the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 8 \) and \( b^2 = 12 \). Therefore, \( a = \sqrt{8} = 2\sqrt{2} \) and \( b = \sqrt{12} = 2\sqrt{3} \). Thus, the length of the semi-major axis is \( 2\sqrt{3} \), and the length of the semi-minor axis is \( 2\sqrt{2} \).

Step 4: Graph the Ellipse

Plot the center of the ellipse at (0, 0). Since the larger value is \( b \), the major axis is along the y-axis. Extend the semi-major axis \( 2 \sqrt{3} \) units up and down from the center, and the semi-minor axis \( 2\sqrt{2} \) units to the left and right of the center. Sketch the ellipse around these points.

Step 5: Find the Foci of the Ellipse

The foci are located along the major axis at a distance \( c \) from the center, where \( c \) is determined by \( c^2 = b^2 - a^2 \). Here, \( b^2 = 12 \) and \( a^2 = 8 \). Thus, \( c = \sqrt{12 - 8} = \sqrt{4} = 2 \). Therefore, the foci of the ellipse are at (0, 2) and (0, -2).

Step 6: Determine the Domain and Range

For the ellipse, the domain is determined by the values \( x \) can take, which range from \(-2\sqrt{2}\) to \(2\sqrt{2}\). The range is determined by the values \( y \) can take, which range from \(-2\sqrt{3}\) to \(2\sqrt{3}\). Thus, the domain is \([-2\sqrt{2}, 2\sqrt{2}]\), and the range is \([-2\sqrt{3}, 2\sqrt{3}]\).

Key Concepts

Standard Form of Ellipse

An ellipse is a fascinating geometric shape, characterized by its oval form. The equation for an ellipse can often be written in a 'standard form,' which helps us easily identify key components like the center, axes, and even the foci. The standard form of an ellipse centered at the origin is given by:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

This equation signifies that the ellipse stretches along the x-axis and y-axis to sizes based on the values of \( a \) and \( b \). Here, \( a \) and \( b \) represent the lengths of the semi-axes, with \( a^2 \) and \( b^2 \) being their corresponding squares. By converting any given ellipse equation into this form, remarkable insights about its structure can be unfolded. For instance, in our example, the initial equation \( 12x^2 + 8y^2 = 96 \) becomes \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \) after simplification. This standard form tells us a lot about the ellipse's measurements.

Ellipse Center

The center of an ellipse is its midpoint, serving as a reference point for locating other features. Its coordinates tell us where the ellipse is situated in the coordinate plane. For ellipses in standard form with no shifts in the equation, like \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \), the center is at the origin, \((0,0)\). This gives symmetry to the ellipse, spreading equally around the origin.

To identify if an ellipse's center is not at the origin, one would typically see additional terms in the equation, shifting \( x \) or \( y \). For example, if the equation had terms like \((x-h)^2\) or \((y-k)^2\), the center would be \((h, k)\). Understanding the location of the center helps significantly in graphing the ellipse and visualizing its placement in space.

Foci of Ellipse

The foci are unique points inside an ellipse that define its shape. They lie along the major axis, which stretches through the widest part of the ellipse. The distance to each focus from the center determines how elongated the ellipse is. To find the foci, we use the relationship:

• \( c^2 = b^2 - a^2 \)

Here, \( c \) is the distance from the center to each focus. In our example, with \( b^2 = 12 \) and \( a^2 = 8 \), we calculate \( c = \sqrt{12 - 8} = 2 \). Therefore, the foci are located at points \((0, 2)\) and \((0, -2)\) along the y-axis.
These points are essential in understanding how the ellipse is proportioned. The closer the foci are to the center, the more circular the ellipse, while further distances create a more stretched look.

Domain and Range of Ellipse

When considering an ellipse, knowing its domain and range is vital for understanding which values \( x \) and \( y \) can take. This information is critical for sketching the graph and understanding its dimensional boundaries. The domain of an ellipse represents the horizontal span of \( x \)-values, while the range shows the vertical span of \( y \)-values.

• Domain: \([-2\sqrt{2}, 2\sqrt{2}]\)
• Range: \([-2\sqrt{3}, 2\sqrt{3}]\)

For our specific ellipse \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \), the domain of \( x \) ranges from \(-2\sqrt{2}\) to \(2\sqrt{2}\), and the range of \( y \) extends from \(-2\sqrt{3}\) to \(2\sqrt{3}\). By examining the domains and ranges, we get a comprehensive picture of the spatial extent of an ellipse.