Problem 25 - A Graphical Approach to College Algebra

Question

Find an equation for each ellipse. \(y\) -intercepts \((0, \pm 2 \sqrt{2}) ;\) foci \((0, \pm 2)\)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{4} + \frac{y^2}{8} = 1\).

Step 1: Identify the Standard Form of the Ellipse

The problem specifies the y-intercepts and foci. Since the y-intercepts are given and the foci have the form (0, ±c), it indicates that the ellipse is vertically oriented. The standard form for a vertically oriented ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\).

Step 2: Use the y-Intercepts to Find 'a'

The y-intercepts of the ellipse are given as \((0, \pm 2\sqrt{2})\), which means \(a = 2\sqrt{2}\). Thus, \(a^2 = (2\sqrt{2})^2 = 8\).

Step 3: Use the Foci to Find 'c'

The foci are given as \((0, \pm 2)\), indicating \(c = 2\).

Step 4: Apply the Relationship Among 'a', 'b', and 'c'

For an ellipse, the relationship is \(c^2 = a^2 - b^2\). Substitute the values found earlier: \(2^2 = 8 - b^2\).

Step 5: Solve for 'b^2'

Start with the equation from step 4: \(4 = 8 - b^2\). Solving for \(b^2\) gives \(b^2 = 4\).

Step 6: Write the Equation of the Ellipse

Now that we have \(a^2 = 8\) and \(b^2 = 4\), substitute these into the standard form \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) to obtain the ellipse's equation: \(\frac{x^2}{4} + \frac{y^2}{8} = 1\).

Key Concepts

Standard Form of an Ellipse

Ellipses are a type of conic section that are defined by their unique oval shape, which can be oriented either horizontally or vertically. Understanding the standard form of the ellipse equation is key to working with them. For a vertically oriented ellipse, the standard form is given by:

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

In this equation, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. When the ellipse is vertical, \(a\) is associated with the y-coordinates, reflecting that the major axis runs up and down. On the other hand, if the major axis were horizontal, the standard form would be:

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

The structure of the formula changes slightly depending on the orientation of the ellipse, helping you identify which axis is longer based on the coefficients.

Foci of an Ellipse

The foci (plural of focus) are two fixed points on the interior of an ellipse used in its formal definition. The distance from the center of the ellipse to a focus is denoted by \(c\). These points are crucial for defining the ellipse because of their geometric properties:

• For any point \(P\) on the ellipse, the sum of the distances to the two foci \(F_1\) and \(F_2\) is constant.

In a vertically oriented ellipse like this one, the foci will be positioned at points \((0, \pm c)\). In the given exercise, the foci are \((0, \pm 2)\), meaning that \(c = 2\). This information also plays a crucial role in understanding how wide the ellipse extends above and below the center.

Y-intercepts of an Ellipse

Intercepts are points where the graph of the ellipse crosses the axes. For ellipses, especially vertically oriented ones, the y-intercepts tell us about the maximum and minimum points on the y-axis. This is pivotal for determining the value of \(a\), the semi-major axis length.

• In this case, the y-intercepts are \((0, \pm 2\sqrt{2})\).

From the y-intercepts, we determine that \(a = 2\sqrt{2}\). When the ellipse meets the y-axis, the whole horizontal component is 0. Thus, the y-intercept values are directly equal to \(a\) and \(-a\). Knowing \(a\) allows us to calculate \(a^2\) by squaring this value, leading to \(a^2 = 8\). This value becomes a part of the ellipse's equation.

Relationship Among a, b, and c

For any ellipse, there's a fundamental relationship between the values of \(a\), \(b\), and \(c\). These variables are linked by the equation:

• \(c^2 = a^2 - b^2\)

This relationship is crucial because it allows you to determine the missing dimension when the other two are known. In our scenario:

• Given that \(c = 2\), \(a = 2\sqrt{2}\), we can find \(b\).

Substitute the known values into the equation, \(4 = 8 - b^2\). Solving this, we get \(b^2 = 4\). This shows us the length of the semi-minor axis squared, completing the parameters needed for our ellipse equation: \(\frac{x^2}{4} + \frac{y^2}{8} = 1\). This connects all components, ensuring the ellipse is properly defined.