Problem 7 - Thomas Calculus

Question

In Exercises \(1-8,\) find the eccentricity of the ellipse. Then find andgraph the ellipse's foci and directrices. $$6 x^{2}+9 y^{2}=54$$

Step-by-Step Solution

Answer

Eccentricity is \(\frac{\sqrt{3}}{3}\); foci at \((\pm \sqrt{3}, 0)\), directrices at \(x = \pm 3\sqrt{3}\).

Step 1: Rewrite the Equation

Start by rewriting the given equation \(6x^{2} + 9y^{2} = 54\) in the form of the standard ellipse equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). Divide the entire equation by 54 to simplify:\[\frac{6x^{2}}{54} + \frac{9y^{2}}{54} = \frac{54}{54}\]Simplify each fraction:\[\frac{x^{2}}{9} + \frac{y^{2}}{6} = 1\]

Step 2: Identify Parameters

Identify the values of \(a^{2}\) and \(b^{2}\) from the simplified equation:\(a^{2} = 9\) so \(a = 3\)\(b^{2} = 6\) so \(b = \sqrt{6}\).

Step 3: Determine Eccentricity

For an ellipse, the eccentricity \(e\) is calculated using the formula \(e = \frac{c}{a}\) where \(c^{2} = a^{2} - b^{2}\).Calculate \(c^{2}\):\[c^{2} = 9 - 6 = 3\]So, \(c = \sqrt{3}\).Now calculate \(e\):\[e = \frac{\sqrt{3}}{3}\]

Step 4: Find the Foci

The foci of the ellipse can be found at \((\pm c, 0)\). With \(c = \sqrt{3}\), the coordinates of the foci are:\((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Step 5: Find the Directrices

The directrices for an ellipse are found using \(x = \pm \frac{a^{2}}{c}\). With \(a^{2} = 9\) and \(c = \sqrt{3}\), compute:\[x = \pm \frac{9}{\sqrt{3}} = \pm 3\sqrt{3}\]

Step 6: Graph the Ellipse and its Elements

Using the standard form equation \(\frac{x^{2}}{9} + \frac{y^{2}}{6} = 1\), plot the ellipse. The major axis is horizontal, with endpoints at \((3,0)\) and \((-3,0)\), and the minor axis is vertical, with endpoints at \((0,\sqrt{6})\) and \((0,-\sqrt{6})\). Plot the foci at \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\) and the directrices \(x = 3\sqrt{3}\) and \(x = -3\sqrt{3}\).

Key Concepts

Standard Form of Ellipse

An ellipse is a geometric shape that can be described using a specific type of equation known as the standard form. The standard form of an ellipse is represented by the equation \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). This form gives us valuable information about the ellipse's orientation and size. In this equation:

• \( a \) is the length of the semi-major axis.
• \( b \) is the length of the semi-minor axis.
• When \( a > b \), the major axis is horizontal. Conversely, if \( b > a \), the major axis is vertical.

To convert any given ellipse equation into this standard form, you need to manipulate it such that the right side of the equation equals 1. For example, the equation \( 6x^{2} + 9y^{2} = 54 \) was simplified by dividing everything by 54, resulting in \( \frac{x^{2}}{9} + \frac{y^{2}}{6} = 1 \). This helps easily identify values for \( a^{2} \) and \( b^{2} \).

Foci of Ellipse

The foci of an ellipse are special points that help define its shape. These points are always located along the major axis, and the sum of distances from any point on the ellipse to the two foci is constant. To find the foci mathematically, we use the value \( c \), which is calculated using the formula \( c^{2} = a^{2} - b^{2} \).

• If \( a^{2} = 9 \) and \( b^{2} = 6 \), then \( c^{2} = 3 \) so \( c = \sqrt{3} \).
• The foci are positioned at \( (\pm c, 0) \) for a horizontal ellipse, or \( (0, \pm c) \) for a vertical ellipse.

Thus, for this example, the foci are positioned at \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \). These points are crucial to the ellipse's definition and can be marked clearly on a graph.

Directrices of Ellipse

The directrices of an ellipse are lines that work with the foci to define the ellipse's eccentricity. The eccentricity \( e \) of an ellipse is calculated with \( e = \frac{c}{a} \). The directrices help in understanding how "stretched" an ellipse is compared to a circle.

• An ellipse with eccentricity close to 0 is more like a circle, while one closer to 1 is more elongated.
• The directrices are defined by the equation \( x = \pm \frac{a^{2}}{c} \) for a horizontal ellipse, or \( y = \pm \frac{a^{2}}{c} \) for a vertical ellipse.

For the given ellipse, we calculate the directrices as \( x = \pm 3\sqrt{3} \). These lines run parallel to the minor axis and help gauge the shape's eccentricity.

Graphing Ellipses

Graphing an ellipse involves plotting its major and minor axes, the foci, and the directrices onto a Cartesian plane. Begin by using the standard form equation to determine the axes:

• The major axis' endpoints are found at \( (\pm a, 0) \) and the minor axis at \( (0, \pm b) \).
• For the example equation \( \frac{x^{2}}{9} + \frac{y^{2}}{6} = 1 \), the major axis endpoints are \( (3, 0) \) and \( (-3, 0) \).
• The minor axis endpoints are \( (0, \sqrt{6}) \) and \( (0, -\sqrt{6}) \).

Then, plot the foci, \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \). Finally, draw the directrices as vertical lines at \( x = 3\sqrt{3} \) and \( x = -3\sqrt{3} \). These elements together create a complete picture of the ellipse's structure, which can now be sketched onto graph paper or a digital plotting tool.