Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Find an equation for each ellipse. Minor axis of length \(4 ;\) foci \((-5,0)\) and \((5,0)\)

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{29} + \frac{y^2}{4} = 1 \).

Step 1: Identify the orientation of the ellipse

Since the foci are along the x-axis at \( (-5, 0) \) and \( (5, 0) \), the major axis is horizontal. Thus, the standard form of the ellipse is \(rac{(x-h)^2}{a^2} + rac{(y-k)^2}{b^2} = 1\).

Step 2: Determine the center of the ellipse

The center of the ellipse is the midpoint between the foci. Calculate the midpoint: \( h = rac{-5 + 5}{2} = 0, \ k = rac{0 + 0}{2} = 0\). Thus, the center is \( (0, 0) \).

Step 3: Calculate the distance between the center and a focus

The distance from the center to each focus \( (c) \) can be calculated as the distance from the center to one of the foci. Here, \( c = 5 \).

Step 4: Determine the length of the minor axis

The minor axis is given as \( 4 \), so \( 2b = 4 \), which means \( b = 2 \).

Step 5: Use relationship between \( a \,b, \,c \) to find \( a \)

In an ellipse, \( c^2 = a^2 - b^2 \.\) Given \( c = 5 \) and \( b = 2\), substitute to find \( a^2: 5^2 = a^2 - 2^2 \). This simplifies to \( 25 = a^2 - 4 \,\ a^2 = 29 \.\)

Step 6: Write the equation of the ellipse

Substitute \( h = 0 \), \( k = 0 \), \( a^2 = 29 \), and \( b^2 = 4 \) into the standard form: \(rac{x^2}{29} + rac{y^2}{4} = 1\). This is the equation of the ellipse.

Key Concepts

Understanding Ellipse Orientation

When determining the orientation of an ellipse, it's important to first look at the placement of the foci. For this problem, the foci are

• at (-5, 0)
• and (5, 0)

This indicates that the foci lie on a horizontal plane along the x-axis. Therefore, the major axis of the ellipse is horizontal. The orientation of the ellipse affects its standard form, which, in the case of a horizontal major axis, can be expressed as \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]In contrast, if the major axis were vertical, the roles of \( a^2 \) and \( b^2 \) in the equation would be swapped.

Locating the Ellipse Center

The center of an ellipse can be found by locating the midpoint between its two foci. This is crucial because the center

• acts as the origin of symmetry for the ellipse
• determines its positioning in the coordinate plane

To compute it for this exercise, we find the midpoint between:

• foci (-5, 0) and (5, 0)

This calculation is straightforward:\[h = \frac{-5 + 5}{2} = 0, \ k = \frac{0 + 0}{2} = 0\]Thus, the center is \((0, 0)\).Identifying this point is essential when crafting the equation of the ellipse, as it serves as the \((h, k)\) values.

Identifying Ellipse Foci

The foci of an ellipse are two fixed points whose distances to any point on the ellipse add up to a constant. These foci are essential in defining the ellipse's shape. In this problem:

• The foci are located at \((-5, 0)\)
• and \((5, 0)\)

The distance from the center to either of the foci is referred to as \(c\). In this particular instance, it calculates as:\[c = \left|5 - 0\right| = 5\]Understanding \(c\) is crucial because it is used in the formula \(c^2 = a^2 - b^2\) which links all significant ellipse parameters.

Determining the Ellipse Axis Lengths

Axes lengths are key characteristics of an ellipse. We have the minor axis length provided directly in the problem statement as 4. The minor axis's semi-length, denoted as \(b\), is thus:

• \(2b = 4 \implies b = 2\)

Now, to find the major axis (semi-major axis length \(a\)), we use the relationship between axes lengths and foci distance given by the formula:\[c^2 = a^2 - b^2\]Substituting the known values:\[5^2 = a^2 - 2^2 \]Simplifying this expression:

• \(25 = a^2 - 4\)
• \(a^2 = 29\)

Having computed \(a^2\) and knowing \(b^2\) along with the center, we can position all these values into the standard ellipse equation:\[\frac{x^2}{29} + \frac{y^2}{4} = 1\]This equation fully represents the ellipse, incorporating all its defining features.