Problem 17 - 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. $$\frac{(x-1)^{2}}{9}+\frac{(y+3)^{2}}{25}=1$$

Step-by-Step Solution

Answer

The center is \((1, -3)\); foci are \((1, 1)\) and \((1, -7)\). Domain: \([-2, 4]\); Range: \([-8, 2]\).

Step 1: Identify the Center

An ellipse equation in standard form is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center of the ellipse. In our equation, \((h, k)\) can be identified as \((1, -3)\).

Step 2: Determine the lengths of the Semi-Major and Semi-Minor Axes

In \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \), compare to the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Notice \( a^2 = 9 \) and \( b^2 = 25 \). Since \( b^2 > a^2 \), the semi-major axis is vertical with length \( b = 5 \), and the semi-minor axis is horizontal with length \( a = 3 \).

Step 3: Find the Foci

For an ellipse, the distance from the center to each focus is given by \( c = \sqrt{b^2 - a^2} \). Hence, \( c = \sqrt{25 - 9} = \sqrt{16} = 4 \). Since the major axis is vertical, the foci are at \((h, k \pm c) = (1, -3 \pm 4)\), or \((1, 1)\) and \((1, -7)\).

Step 4: Determine the Domain and Range

The domain and range are determined by the extents of the ellipse along the x and y axes. The ellipse centered at \((1, -3)\) has a horizontal extent spanning from \(x = 1 - a\) to \(x = 1 + a\), and a vertical extent from \(y = -3 - b\) to \(y = -3 + b\). Thus, the domain is \([-2, 4]\) and the range is \([-8, 2]\).

Step 5: Graph the Ellipse

Using the center \((1, -3)\), plot this point. From there, measure 3 units left and right for the endpoints of the semi-minor axis, to \((-2, -3)\) and \((4, -3)\). For the vertical semi-major axis, measure 5 units up and down, to \((1, 2)\) and \((1, -8)\). Plot the foci \((1, 1)\) and \((1, -7)\). Draw a smooth oval shape connecting these points to represent the ellipse.

Key Concepts

Graphing Ellipses

When graphing an ellipse, it's essential first to identify key points and dimensions that will shape the curve. Every ellipse has a center, semi-major and semi-minor axes, and foci that guide its layout.
Start by locating the center of the ellipse. For the given equation \(\frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1\), the center is at \((1, -3)\). This point serves as the midpoint and balances the ellipse.
Next, identify the lengths of the semi-major and semi-minor axes. Since \(b^2 = 25\) is greater than \(a^2 = 9\), the longer, semi-major axis is vertical, measuring \(5\) units. Conversely, the semi-minor axis is horizontal, stretching \(3\) units.

• Plot the ends of the semi-minor axis by moving 3 units left to \((-2, -3)\) and right to \((4, -3)\).
• For the semi-major axis, mark 5 units upwards to \((1, 2)\) and downwards to \((1, -8)\).

Finally, connect these boundary points with a smooth, oval-like shape, maintaining symmetry around the center, for your accurate depiction of the ellipse.

Equation of Ellipse

Understanding the equation of an ellipse is foundational for graphing and identifying its properties. The equation \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \) is a standard form:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where

• \( (h, k) \) represents the center.
• \( a^2 \) and \( b^2 \) denote the denominators under \(x\) and \(y\) terms, respectively.

For our equation, \(h = 1\) and \(k = -3\), positioning the center at \((1, -3)\). The values \(a^2 = 9\) and \(b^2 = 25\) give axis lengths \(a = 3\) and \(b = 5\).
Since \(b^2\) exceeds \(a^2\), this ellipse is longer in the vertical direction. Such properties are crucial for graphing and calculating further attributes like the foci or domain and range.

Foci of Ellipse

The foci are two special points inside an ellipse that help define its shape. They lie along the major axis and are key in understanding how 'stretched' the ellipse is.
To find the foci, calculate the distance \(c\) from the center to each focus using the formula:
\[ c = \sqrt{b^2 - a^2} \]
Insert the values from \(b^2 = 25\) and \(a^2 = 9\), obtaining \(c = \sqrt{16} = 4\).

• Since the major axis is vertical, place the foci at \((h, k \pm c) = (1, -3 \pm 4)\)
• This results in foci located at \((1, 1)\) and \((1, -7)\).

These points are just inside the curve vertically, aligning closely with the ellipse’s shape and keeping it uniformly stretched along the vertical axis.

Domain and Range

The domain and range of an ellipse encapsulate the set of all possible x and y values the ellipse can take. This is critical in understanding the ellipse's placement on the Cartesian plane.
From the equation \(\frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1\) and knowing the center is \((1, -3)\), determine the domain and range:

• The horizontal domain extends 3 units outward from the center point, resulting in \([-2, 4]\).
• Vertically, it spans 5 units from the center, providing a range of \([-8, 2]\).

Both domain and range highlight the ellipse's coverage on the x-axis and y-axis, signifying its width and height. This helps in precisely locating and drawing the ellipse in its correct position.