Problem 8 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$4 x^{2}+25 y^{2}=100$$

Step-by-Step Solution

Answer

Vertices: (5, 0), (-5, 0); Foci: (±√21, 0); Eccentricity: 0.91; Major Axis: 10; Minor Axis: 4.

Step 1: Write the equation in standard form

Rewrite the given ellipse equation \(4x^2 + 25y^2 = 100\) by dividing all terms by 100. This will help us identify the standard form of the ellipse equation. \[ \frac{x^2}{25} + \frac{y^2}{4} = 1 \] This is in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a^2 = 25\) and \(b^2 = 4\).

Step 2: Identify the major and minor axes

In the equation \(\frac{x^2}{25} + \frac{y^2}{4} = 1\), we see that \(25 > 4\). Hence, \(a^2 = 25\) and \(b^2 = 4\), indicating the ellipse is elongated along the x-axis. Thus, the major axis is the x-axis and the minor axis is the y-axis.

Step 3: Compute the vertices

The vertices along the major x-axis are given by \((\pm a, 0)\). Since \(a^2 = 25\), \(a = 5\). Therefore, the vertices are \((5, 0)\) and \((-5, 0)\).

Step 4: Compute the foci

The foci of an ellipse are given by \((\pm c, 0)\) for an ellipse elongated along the x-axis, where \(c = \sqrt{a^2 - b^2}\). Using \(a^2 = 25\) and \(b^2 = 4\), we have:\[ c = \sqrt{25 - 4} = \sqrt{21} \approx 4.58 \]Thus, the foci are \((\sqrt{21}, 0)\) and \((-\sqrt{21}, 0)\).

Step 5: Determine the eccentricity

The eccentricity \(e\) of an ellipse is defined as \(e = \frac{c}{a}\), where \(c = \sqrt{21}\) and \(a = 5\). Calculate the eccentricity:\[ e = \frac{\sqrt{21}}{5} \approx 0.91 \]

Step 6: Determine lengths of the major and minor axes

The length of the major axis is \(2a\) while the length of the minor axis is \(2b\). Hence,- Major Axis: \(2 \times 5 = 10\)- Minor Axis: \(2 \times 2 = 4\)

Step 7: Sketch the ellipse

Sketch the ellipse centered at the origin \((0,0)\), with vertices \((5,0)\), \((-5,0)\), minor axis endpoints \((0,2)\) and \((0,-2)\), and foci at approximately \((\pm4.58, 0)\). It should appear elongated along the x-axis.

Key Concepts

Vertices

In the context of an ellipse, vertices are key points that define its shape. Since our given ellipse is defined by the equation \[ \frac{x^2}{25} + \frac{y^2}{4} = 1 \], the vertices lie along the lengthiest part of the ellipse, known as the major axis. For our specific ellipse, the vertices are calculated by determining \(a\), which is the value derived from the equation \(a^2 = 25\), giving us \(a = 5\). The vertices are therefore located at

• \((5, 0)\) and
• \((-5, 0)\)

These points are critical because they relate directly to the maximum width of the ellipse when extended through its center point along the x-axis, depicting how the ellipse is elongated horizontally. Knowing the placement of the vertices helps in accurately sketching the ellipse's overall shape.

Foci

The foci of an ellipse are two specific points that reside inside the ellipse. They hold unique importance, contributing to the geometric property of the ellipse that the sum of the distances from any point on the ellipse to the two foci is constant. For the ellipse in question, the foci are determined using the formula \(c = \sqrt{a^2 - b^2}\). Given \(a^2 = 25\) and \(b^2 = 4\), we calculate:

• \(c = \sqrt{25 - 4} = \sqrt{21} \approx 4.58\)

Thus, the foci are located at

• \((\sqrt{21}, 0)\)
• \((-\sqrt{21}, 0)\)

These points are crucial for understanding the shape dynamics of the ellipse and are positioned along the major axis, suggesting their influence on the ellipse's overall configuration.

Eccentricity

Eccentricity in ellipses is a measure that describes how much the shape deviates from being circular. Represented by \(e\), the eccentricity gives insights into the shape's elongation. The formula used to calculate the eccentricity of an ellipse is \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus, and \(a\) is the distance to a vertex. In our solution:

• \(c = \sqrt{21}\)
• \(a = 5\)

The eccentricity is calculated as:

• \(e = \frac{\sqrt{21}}{5} \approx 0.91\)

An eccentricity of \(0.91\) indicates a significantly elongated ellipse, implying that despite being stretched along the x-axis, the ellipse retains its distinct shape without collapsing into a circle or a line.

Major Axis

The major axis of an ellipse is the longest diameter that passes through its center and both foci. It represents the most extended part of the ellipse and dictates the direction in which the ellipse is stretched. For the provided ellipse equation \(\frac{x^2}{25} + \frac{y^2}{4} = 1\), the major axis is the x-axis, due to \(a^2\) being greater than \(b^2\). The length of the major axis is calculated as \(2a\), where \(a = 5\):

• Major Axis Length: \(2 \times 5 = 10\)

Understanding this helps visualize the ellipse's overall orientation and size, confirming that it is horizontally extended and centered at the origin \((0,0)\). The insights from the major axis length are pivotal for accurately drawing the ellipse, as it heavily bounds the vertex points.

Minor Axis

The minor axis of an ellipse is the shortest diameter, lying perpendicular to the major axis, and similarly passing through the center. Even though it is shorter, it plays an integral role in finalizing the ellipse's width. For our ellipse equation \(\frac{x^2}{25} + \frac{y^2}{4} = 1\), the minor axis lies along the y-axis, acknowledged because \(b^2 = 4\) is less than \(a^2 = 25\). The length of the minor axis is calculated using the formula \(2b\), where \(b = 2\):

• Minor Axis Length: \(2 \times 2 = 4\)

This dimension determines the compactness of the ellipse along its vertical, influencing the shape's overall profile when combined with the major axis. Observing the minor axis helps provide a clearer picture of the ellipse's full form.