Problem 17 - A Graphical Approach to College Algebra

Question

Graph each ellipse by hand. Give the domain and range. Give the foci and identify the center. Do not use a calculator. $$\frac{(x-1)^{2}}{9}+\frac{(y+3)^{2}}{25}=1$$

Step-by-Step Solution

Answer

Center: (1, -3), Foci: (1, 1) and (1, -7), Domain: [-2, 4], Range: [-8, 2].

Step 1: Identify the Standard Form

Recognize that the equation \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \) represents an ellipse in standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center.

Step 2: Determine the Center

From the equation \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \), identify the center of the ellipse as \((h, k) = (1, -3)\).

Step 3: Identify the Axes Lengths

Here \(a^2 = 9\) (horizontal) and \(b^2 = 25\) (vertical), meaning \(a = 3\) and \(b = 5\). The b value is larger, indicating the major axis is vertical.

Step 4: Determine the Foci

Use the formula \(c^2 = b^2 - a^2\) to find \(c\). Calculate \(c^2 = 25 - 9 = 16\), so \(c = 4\). Hence, the foci are \((1, -3 + 4) = (1, 1)\) and \((1, -3 - 4) = (1, -7)\).

Step 5: Domain and Range Identification

The domain is determined by the horizontal extent of the ellipse: \([1 - 3, 1 + 3] = [-2, 4]\). The range is determined by the vertical extent: \([-3 - 5, -3 + 5] = [-8, 2]\).

Step 6: Sketch the Ellipse

Draw the ellipse centered at \((1, -3)\), stretching 3 units left and right (horizontal distance) and 5 units up and down (vertical distance). Mark the center, vertices at \((1, 2)\) and \((1, -8)\), and the foci at \((1, 1)\) and \((1, -7)\).

Key Concepts

Standard Form of an Ellipse

An ellipse can often be represented in a very specific way called the "standard form." The standard form of an ellipse is useful because it reveals important characteristics at a glance. For an ellipse, the standard form equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here,

• \((h, k)\) is the center of the ellipse.
• \(a^2\) and \(b^2\) indicate the squared lengths of the semi-axes.

If \(a^2\) is greater than \(b^2\), the ellipse is horizontal. If \(b^2\) is greater than \(a^2\), then the ellipse is vertical. The presence of a "1" on the right side of the equation is a hallmark of this form. This critical structure makes it easy to find other properties of the ellipse, like its center, the length of its axes, and the orientation. Knowing this form allows a quick identification of these properties from the equation itself.
In our example, \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \) fits perfectly into this form, ready to tell us much about the ellipse's dimensions and position.

Center of an Ellipse

The center of an ellipse is the point around which the entire ellipse is evenly distributed. It's symbolized by the coordinates \((h, k)\), which you find in the standard form equation. This center is like the heart of the ellipse, a place equidistant from all the critical points on the shape.
Given an equation in standard form, such as \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \), you can easily pick out the center by identifying the values of \(h\) and \(k\). For this equation, \(h = 1\) and \(k = -3\), so the center is at the point \((1, -3)\).
Placing the center on your graph is the first step in sketching your ellipse, setting the foundation for determining directions and distances to the vertices and foci.

Foci of an Ellipse

The foci of an ellipse are two distinct points inside the ellipse that have a special property: the sum of the distances from these points to any point on the ellipse is constant. This property is what gives the ellipse its unique shape.
Finding the foci involves a little calculation. You start with the formula \(c^2 = b^2 - a^2\), derived from the standard form parameters.

• In our equation \( \frac{(x-1)^2}{9} + \frac{(y+3)^2}{25} = 1 \), we have \(a^2 = 9\) and \(b^2 = 25\), hence \(c^2 = 25 - 9 = 16\).
• This gives \(c = 4\), which tells us how far the foci are from the center along the major axis.
• Adding and subtracting \(c\) from the center's y-coordinate \((-3)\), we find our foci at \((1, 1)\) and \((1, -7)\).

The foci are quite important as they help in understanding the ellipse's geometry and are especially crucial in scientific applications like astronomy.

Domain and Range

In mathematics, when we discuss ellipses, the terms "domain" and "range" refer to the set of possible values that the ellipse can have in the x-axis and y-axis directions, respectively. They describe the extent of the ellipse along these axes.
The domain is the horizontal span of the ellipse. For our standard form equation,

• Identify the horizontal distance \(a\), which is 3 units from the center \((1, -3)\) in our example.
• Thus, the domain from left to right is \([1 - 3, 1 + 3] = [-2, 4]\).

Similarly, the range is the vertical span:

• Here, \(b = 5\), so the ellipse stretches 5 units vertically from the center.
• Hence, the range is expressed as \([-3 - 5, -3 + 5] = [-8, 2]\).

Knowing the domain and range helps in plotting the ellipse correctly on a coordinate plane and understanding its reach in all directions.