Problem 16 - Precalculus: Mathematics for Calculus

Question

An equation of an ellipse is given. (a) Find the vertices, foci, and eccentricity of the ellipse. (b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse. $$4 x^{2}+25 y^{2}=100$$

Step-by-Step Solution

Answer

Vertices: \((-5, 0)\) and \((5, 0)\), Foci: \((\pm \sqrt{21}, 0)\), Eccentricity: \(\frac{\sqrt{21}}{5}\). Major axis: 10, Minor axis: 4.

Step 1: Write the Standard Form of the Ellipse

Start with the equation of the ellipse: \( 4x^2 + 25y^2 = 100 \). Divide both sides by 100 to express it in the standard form of an ellipse: \( \frac{x^2}{25} + \frac{y^2}{4} = 1 \). Recognize this as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with \( a^2 = 25 \) and \( b^2 = 4 \).

Step 2: Determine the Vertices

Since \( a^2 = 25 \), we have \( a = 5 \). Similarly, \( b^2 = 4 \), so \( b = 2 \). The vertices are located at \( (\pm a, 0) \), so the vertices are \((-5, 0)\) and \((5, 0)\).

Step 3: Calculate the Foci

The formula for the distance from the center to the foci of an ellipse is \( c = \sqrt{a^2 - b^2} \). So, \( c = \sqrt{25 - 4} = \sqrt{21} \). Thus, the foci are at \((\pm \sqrt{21}, 0)\).

Step 4: Find the Eccentricity

The eccentricity \( e \) of an ellipse is given by \( e = \frac{c}{a} \). Substituting the values: \( e = \frac{\sqrt{21}}{5} \).

Step 5: Determine the Lengths of Major and Minor Axes

The major axis has a length of \( 2a \), which is \( 2 \times 5 = 10 \). The minor axis has a length of \( 2b \), which is \( 2 \times 2 = 4 \).

Step 6: Sketch the Graph of the Ellipse

Draw the ellipse centered at the origin (0,0). Mark the vertices \((5,0)\) and \((-5,0)\) along the x-axis. Mark the ends of the minor axis \((0,2)\) and \((0,-2)\) along the y-axis. Sketch an oval shape that passes through these four points.

Key Concepts

Vertices of Ellipse

The vertices of an ellipse are important points that determine its overall shape. When dealing with the ellipse equation in standard form, such as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the vertices are found on the axes that align with the larger of the two denominators, \(a^2\) or \(b^2\). In the given problem, we have the standard form equation \( \frac{x^2}{25} + \frac{y^2}{4} = 1 \).
- Here, \( a^2 = 25 \) and \( b^2 = 4 \), indicating the major axis is along the x-axis.- Calculate \( a \) by taking the square root of \( a^2 \), which gives \( a = 5 \).- The vertices, therefore, are located at points \((\pm a, 0)\), or \((-5, 0)\) and \((5, 0)\).
These points are crucial because they define the longest width of the ellipse, also known as the major axis. This is why the vertices are always equidistant from the center.

Foci of Ellipse

Foci (or focuses) of an ellipse are pivotal points that lie on the ellipse's major axis. They are used to construct an ellipse based on the concept that the sum of the distances from any point on the ellipse to the two foci is constant. This unique property distinguishes ellipses from other conic sections.
To find the foci:- Use the formula \( c = \sqrt{a^2 - b^2} \) which helps us determine their location.- From the previous exercise, we have \( a^2 = 25 \) and \( b^2 = 4 \). Plug these into the formula: \( c = \sqrt{25 - 4} = \sqrt{21} \).- Therefore, the foci are positioned at \((\pm \sqrt{21}, 0)\), along the x-axis, since it is the major axis.
The foci allow us to understand the elongation of the ellipse. The closer the foci are to each other, the more circular the ellipse. Conversely, the further apart they are, the more the ellipse resembles a stretched oval.

Eccentricity of Ellipse

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. It’s a non-negative number defined for every ellipse, visually representing how stretched or squished the shape appears. Mathematically, eccentricity \( e \) is calculated using the formula \( e = \frac{c}{a} \), where \( c \) is the focal distance from the center to a focus.
In our example:- We previously found \( c = \sqrt{21} \) and \( a = 5 \).- Thus, the eccentricity is \( e = \frac{\sqrt{21}}{5} \).
The result is a number between 0 and 1:- If \( e = 0 \), the shape is a circle.- If \( 0 < e < 1 \), it’s an ellipse, which is the case here.
Eccentricity helps to quickly assess how stretched an ellipse is just by a glance at its value.

Major and Minor Axes of Ellipse

The axes of an ellipse define its dimensions and orientation. The major and minor axes are right-angled diameters that intersect at the center of the ellipse. Knowing their lengths provides insight into the size and shape of the ellipse.
For the equation \( \frac{x^2}{25} + \frac{y^2}{4} = 1 \):- The **major axis** corresponds to the largest value in the denominator (\( a^2 = 25 \)), aligned with the x-axis in this instance. Its length is \( 2a \), calculated as \( 2 \times 5 = 10 \).- The **minor axis** relates to the smaller value (\( b^2 = 4 \)), aligned with the y-axis. Its length is \( 2b \), ensuring \( 2 \times 2 = 4 \).
These axes allow an easy visual representation of the ellipse size. The major axis is always the longer one, stretching through the ellipse's center and aligned parallel with the vertices.