Problem 18 - Precalculus Mathematics 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. $$9 x^{2}+4 y^{2}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±1/2), Foci: (0, ±√5/6), Eccentricity: √5/3, Major axis length: 1, Minor axis length: 2/3.

Step 1: Standard Form of the Ellipse Equation

First, divide all terms in the given equation by 1 to have the standard form of the ellipse. However, since it is already equal to 1, identify the coefficients to rewrite the equation in standard form \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]. The given equation is \[ 9x^2 + 4y^2 = 1 \]. Divide everything by 1:\[ \frac{x^2}{1/9} + \frac{y^2}{1/4} = 1 \]. This gives us:\[ \frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1 \].

Step 2: Identifying Major and Minor Axes

The standard form equation is \[ \frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1 \]. Since \((1/2)^2 > (1/3)^2\), the major axis is along the y-axis. The lengths of the semi-major and semi-minor axes are \(b = 1/2\) and \(a = 1/3\), respectively.

Step 3: Find the Vertices

Vertices occur at the ends of the major axis. Since the major axis is vertical, the vertices are at: \((0, \pm b) = (0, \pm 1/2)\)

Step 4: Find the Foci

The foci of the ellipse are given by \((0, \pm c)\). To find \(c\), use \[c^2 = b^2 - a^2\]. Compute \[c^2 = (1/2)^2 - (1/3)^2 = 1/4 - 1/9\]. Find a common denominator and calculate: \[ c^2 = 9/36 - 4/36 = 5/36\]. Thus, \[c = \sqrt{5}/6\].Therefore, the foci are at \((0, \pm \sqrt{5}/6)\).

Step 5: Calculate the Eccentricity

The eccentricity of an ellipse is given by \[ e = \frac{c}{b} \]. Here, \[e = \frac{\sqrt{5}/6}{1/2} = \frac{\sqrt{5}}{3}\].

Step 6: Determine the Lengths of the Major and Minor Axes

The length of the major axis is \[2b = 2(1/2) = 1\].The length of the minor axis is \[2a = 2(1/3) = 2/3\].

Step 7: Sketch the Ellipse

On a graph, plot the vertices at \((0, \pm 1/2)\). The ellipse will be elongated along the y-axis due to its longer major axis. Mark the foci within the ellipse at \((0, \pm \sqrt{5}/6)\). Use these points to sketch a symmetrical ellipse centered at the origin.

Key Concepts

Vertices

In the context of an ellipse, vertices are crucial as they define the ends of the longest diameter. For any ellipse, the vertices are located at the endpoints of its major axis.
For an equation in the standard form \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \], the vertices are given by:

• If \( b^2 > a^2 \), as in our case, the major axis is vertical, and the vertices are at \((0, \pm b)\).
• If \( a^2 > b^2 \), the major axis is horizontal, and the vertices are at \((\pm a, 0)\).

In this problem, since our equation is \( \frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1 \), the vertices are \((0, \pm 1/2)\). These points are important because they give a clear direction and outline of the elliptical shape.

Foci

The foci, pronounced "foh-sigh," are two special points that lie along the major axis of the ellipse. They are key in maintaining the shape of an ellipse as their presence ensures that the sum of the distances from any point on the ellipse to the foci remains constant. For an ellipse, the formula to determine the foci is:

• \( c^2 = b^2 - a^2 \)
• The foci are at \((0, \pm c)\) for a vertical major axis, and \((\pm c, 0)\) for a horizontal major axis

In our example, we calculated \( c = \sqrt{5}/6 \), so the foci are located at \((0, \pm \sqrt{5}/6)\). The foci are situated inside the ellipse, between its center and each vertex.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a circle. It ranges from 0 to less than 1, where 0 is a perfect circle, and values closer to 1 indicate a more elongated ellipse.To find the eccentricity \( e \) of an ellipse, use the formula:

• \( e = \frac{c}{b} \)

From the problem, we established \( c = \sqrt{5}/6 \) and \( b = 1/2 \), resulting in an eccentricity of \( e = \frac{\sqrt{5}}{3} \). This number is relatively small, indicating that while the ellipse is elongated, it is not extremely stretched.

Major axis

The major axis of an ellipse is the longest diameter that runs through the center, spanning from one vertex to the opposite vertex. For a standard form equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the length of the major axis is determined by the larger value of \( a \) or \( b \):

• If \( b^2 > a^2 \), the major axis is vertical, and its length is \( 2b \).
• If \( a^2 > b^2 \), the major axis is horizontal, and its length is \( 2a \).

Here, \( b = 1/2 \), so the length of the major axis is \( 1 \). This axis runs along the y-axis, centering the ellipse vertically.

Minor axis

The minor axis is the shortest diameter of the ellipse and is perpendicular to the major axis. It also passes through the center of the ellipse but is shorter than the major axis.In the ellipse's standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the minor axis is determined by:

• If \( b^2 > a^2 \), the minor axis is horizontal, and its length is \( 2a \).
• If \( a^2 > b^2 \), the minor axis is vertical, and its length is \( 2b \).

For our equation, since \( a = 1/3 \), the length of the minor axis is \( 2/3 \). This short axis extends horizontally through the ellipse's center.