Problem 38 - College Algebra with Modeling and Visualization

Question

Sketch a graph of the ellipse. $$ x^{2}+\frac{(y-3)^{2}}{4}=1 $$

Step-by-Step Solution

Answer

The ellipse center is (0, 3) with a vertical semi-major axis of 2 and a horizontal semi-minor axis of 1.

Step 1: Identify the Standard Form of the Ellipse

The given equation of the ellipse is \( x^2 + \frac{(y-3)^2}{4} = 1 \). The standard form of an ellipse centered at \((h, k)\) with a semi-major axis \(a\) and semi-minor axis \(b\) is \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) or \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Our equation is already in this form.

Step 2: Determine the Center of the Ellipse

From the equation \( x^2 + \frac{(y-3)^2}{4} = 1 \), we can see it's of the form \( (x-h)^2 \) and \( \frac{(y-k)^2}{4} \). Thus, the center of the ellipse is \((h, k) = (0, 3)\).

Step 3: Identify the Axes and Lengths

The term \(x^2\) corresponds to \( \frac{(x-0)^2}{1} \), indicating \(b^2=1\) and \(b=1\). The term \(\frac{(y-3)^2}{4}\) indicates \(a^2=4\) and \(a=2\). Since the larger denominator is under \((y-3)^2\), the semi-major axis is vertical.

Step 4: Plot the Center and Axes

First, plot the center of the ellipse at \((0, 3)\). Your semi-major axis (vertical) will be 2 units above and below the center (\(0, 5\) and \(0, 1\)), and the semi-minor axis (horizontal) will be 1 unit left and right of the center (\((1, 3)\) and \((-1, 3)\)).

Step 5: Sketch the Ellipse

Using the plotted points from Step 4, draw a smooth, oval shape connecting the ends of the axes. Ensure the shape is centered at \((0, 3)\) and touches the plotted points on the axes.

Key Concepts

Standard Form of Ellipse

An ellipse is one of the most common conic sections you will encounter, and describing it using its standard form is crucial for understanding its properties. The standard form of an ellipse equation is generally written as:

• \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• Or similarly, \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\).

Here, \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. The position of \(a\) and \(b\) will determine the orientation of the ellipse, whether it lies horizontally or vertically.
In the given exercise, the equation \(x^2 + \frac{(y-3)^2}{4} = 1\) matches the standard form where \(h = 0\), \(k = 3\), and hence, requires no additional manipulation to identify its properties.

Center of Ellipse

The center of an ellipse is the point around which the ellipse is symmetrically arranged. In the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the center is denoted by the coordinates \((h, k)\).
For our specific equation \(x^2 + \frac{(y-3)^2}{4} = 1\), the center is found by identifying the transformations applied in the equation:

• The term \(x^2\) can be seen as \((x-0)^2\), so \(h = 0\).
• The term \((y-3)^2\) gives \(k = 3\).

Thus, the center of the ellipse is located at \((0, 3)\). This point is crucial as it helps in plotting the ellipse accurately on a coordinate plane.

Semi-major and Semi-minor Axis

The semi-major and semi-minor axes determine the size and shape of the ellipse. They are half of the major and minor axes, the longest and shortest diameters, respectively. For identifying these from the standard form, observe the denominators in the equation.

• If \(a^2 > b^2\), the semi-major axis will be aligned according to \(a\), and if \(b^2 > a^2\), vice versa.

In our problem’s equation, \(x^2 + \frac{(y-3)^2}{4} = 1\),

• The term \(x^2\) acts as \(\frac{(x-0)^2}{1}\), showing that \(b^2 = 1\) (hence \(b=1\)).
• The term \(\frac{(y-3)^2}{4}\) establishes \(a^2 = 4\) (hence \(a=2\)).

Since \(a^2 = 4\) is greater than \(b^2 = 1\), the major axis is vertical. With the center at \((0, 3)\), the ellipse extends 2 units up and down (to \(0, 1\) and \(0, 5\)) and 1 unit left to right (to \(1, 3\) and \(-1, 3\)). This configuration is crucial when sketching the ellipse and understanding its alignment.