Problem 20 - 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. $$\frac{(x+3)^{2}}{25}+\frac{(y+2)^{2}}{36}=1$$

Step-by-Step Solution

Answer

Center: (-3, -2); Domain: [-8, 2]; Range: [-8, 4]; Foci: (-3, -2±√11).

Step 1: Identify the Center

The standard form of an ellipse is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center of the ellipse. Given the equation, \(\frac{(x+3)^2}{25} + \frac{(y+2)^2}{36} = 1\), the center is \((-3, -2)\).

Step 2: Determine Foci and Axes

For ellipses, if \( b^2 > a^2 \), the major axis is vertical. Here, \( a^2 = 25 \) and \( b^2 = 36 \), so the major axis is vertical. The foci are given by \( c = \sqrt{b^2 - a^2} = \sqrt{36 - 25} = \sqrt{11} \). They are located at \((-3, -2 \pm \sqrt{11})\).

Step 3: Find Domain and Range

Since the center is at \((-3, -2)\), with a semi-major axis (vertically) of 6 and a semi-minor axis (horizontally) of 5, the domain is from \(-8\) to \(2\): \([-8, 2]\), and the range is from \(-8\) to \(4\): \([-8, 4]\).

Step 4: Sketch the Ellipse

To sketch, plot the center at \((-3, -2)\), then mark points 5 units left and right (\((-8, -2)\) and \((2, -2)\)) and 6 units up and down (\((-3, 4)\) and \((-3, -8)\)). Draw the elliptical shape using these points as guides.

Key Concepts

Graphing

Graphing an ellipse involves plotting points in a specific way to form the elongated circular shape. Begin by identifying the center of the ellipse, which serves as the reference point for plotting. In our case, the center is at \((-3, -2)\). From here, using the semi-major and semi-minor axes derived from the denominators in the ellipse equation, locate additional points.

• The semi-major axis is determined by the larger denominator, \(b^2 = 36\), suggesting a vertical span of 6 units. This translates to plotting 6 units above and 6 units below the center.
• The semi-minor axis corresponds to the smaller denominator, \(a^2 = 25\). This gives a horizontal spread of 5 units, plotted 5 units to the left and 5 units to the right of the center.

Connect these plotted points in a smooth, oval shape around the center.

Domain and Range

Understanding the domain and range of an ellipse involves recognizing its horizontal and vertical spread. The domain pertains to the possible \(x\)-values that the ellipse covers, while the range involves the potential \(y\)-values.
For our ellipse, the horizontal spread (domain) is influenced by the semi-minor axis:

• Start at the center, \((-3, -2)\), and measure 5 units left to \(-8\) and 5 units right to \(2\).
• Thus, the domain is \([-8, 2]\).

The vertical spread (range) is controlled by the semi-major axis:

• Starting from the center, extend 6 units up to \(4\) and 6 units down to \(-8\).
• This makes the range \([-8, 4]\).

Foci

The foci of an ellipse are vital for defining its shape. They lie along the major axis, equidistant from the center. To find these, use the relationship \(c = \sqrt{b^2 - a^2}\).
Given our parameters, compute as follows:

• \(a^2 = 25\) and \(b^2 = 36\), with \(b^2\) being greater, confirms a vertical major axis.
• Calculate \(c = \sqrt{36 - 25} = \sqrt{11}\).
• Position the foci at \((-3, -2 \pm \sqrt{11})\).

This placement along the vertical axis indicates the ellipse's most stretched direction, a key characteristic in the graph.

Center of Ellipse

The center of an ellipse signifies the balancing point from which the figure expands. It is determined from the general ellipse equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) denotes the center.
From our specific equation, \(\frac{(x+3)^2}{25} + \frac{(y+2)^2}{36} = 1\):

• Observe that \(h = -3\) and \(k = -2\).
• Thus, the center is \((-3, -2)\).

Understanding the center is crucial, not just for graphing, but also for determining the ellipse's symmetry and orientation.