Problem 5 - Thomas Calculus

Question

In Exercises \(1-8,\) find the eccentricity of the ellipse. Then find and graph the ellipse's fociand directrices. $$ 3 x^{2}+2 y^{2}=6 $$

Step-by-Step Solution

Answer

Eccentricity is \(\approx 0.577\); foci at \((0, \pm 1)\); directrices at \(y = \pm 3\).

Step 1: Convert Equation to Standard Form

First, we need to express the given equation \(3x^2 + 2y^2 = 6\) in the standard form of an ellipse. Divide the entire equation by 6:\[ \frac{3x^2}{6} + \frac{2y^2}{6} = 1 \]Simplify the fractions:\[ \frac{x^2}{2} + \frac{y^2}{3} = 1 \]This equation is now in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a^2 = 2\) and \(b^2 = 3\).

Step 2: Identify Orientation and Parameters

Since \(b^2 > a^2\), the ellipse is vertically oriented. From the standard form, we identify:\[ a^2 = 2 \quad \Rightarrow \quad a = \sqrt{2} \approx 1.41 \]\[ b^2 = 3 \quad \Rightarrow \quad b = \sqrt{3} \approx 1.73 \]

Step 3: Calculate Eccentricity

The eccentricity \(e\) of an ellipse is calculated using the formula for vertical ellipses:\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]Substituting the values \(a^2 = 2\) and \(b^2 = 3\):\[ e = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} \approx 0.577 \]

Step 4: Determine Coordinates of Foci

The foci of a vertical ellipse are located at \((0, \pm c)\), where \(c = be\).Calculate \(c\):\[ c = b \times e = \sqrt{3} \times \sqrt{\frac{1}{3}} = 1 \]Thus, the foci are at \((0, \pm 1)\).

Step 5: Find Equations of Directrices

The equations for the directrices of a vertical ellipse are at \(y = \pm \frac{b^2}{c}\).Calculate the directrices:\[ y = \pm \frac{b^2}{c} = \pm \frac{3}{1} = \pm 3 \]So, the directrices are at \(y = 3\) and \(y = -3\).

Step 6: Graph the Ellipse, Foci, and Directrices

Draw the ellipse centered at the origin with the axes aligned. Plot the vertices at \((0, \pm \sqrt{3})\) and \((\pm \sqrt{2}, 0)\). Mark the foci at \((0, 1)\) and \((0, -1)\). Draw the horizontal lines for the directrices at \(y = 3\) and \(y = -3\).

Key Concepts

Standard Form of an Ellipse

To solve problems involving ellipses, it's important to start with the equation in its standard form. The standard form of an ellipse equation is given as \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\]where:

• \( a^2 \) is the denominator under the \( x^2 \) term.
• \( b^2 \) is the denominator under the \( y^2 \) term.
• When the equation equals 1, it signifies that the ellipse is properly scaled.
• If \( a^2 < b^2 \), this indicates that the ellipse is vertically oriented.
• If \( a^2 > b^2 \), it is horizontally oriented.

The given exercise starts with a non-standard form: \[ 3x^2 + 2y^2 = 6. \]To convert it to standard form, we divide each term by 6:\[ \frac{x^2}{2} + \frac{y^2}{3} = 1. \]This equation now fits the standard form, allowing us to easily determine other properties of the ellipse.

Vertical Ellipse Orientation

Once the equation is in standard form, the orientation of the ellipse can be identified by comparing \( a^2 \) and \( b^2 \).

• In this exercise, the equation \( \frac{x^2}{2} + \frac{y^2}{3} = 1 \) has \( a^2 = 2 \) and \( b^2 = 3 \).
• Since \( b^2 > a^2 \), the ellipse is vertically oriented.
• The major axis aligns with the y-axis, meaning the endpoints (vertices) of the longer axis are along this axis.

Identifying the orientation is crucial as it affects how other properties like foci and directrices are calculated and graphically represented.

Ellipse Foci and Directrices

The foci and directrices are key features that further define an ellipse's shape.

• The foci (\( F \)), two fixed points inside the ellipse, are located along its major axis.
• For a vertical ellipse, they are at points \( (0, \pm c) \), where \( c = b imes e \) (with \( e \) being the eccentricity).

The eccentricity (\( e \)) measures the deviation of the ellipse from being a circle, calculated using:\[e = \sqrt{1 - \frac{a^2}{b^2}}. \]Substituting the values \( a^2 = 2 \) and \( b^2 = 3 \) yields:\[e = \sqrt{\frac{1}{3}} \approx 0.577.\]Then, calculate \( c \):\[ c = 1. \]Thus, the foci are at \( (0, \pm 1) \).The directrices are vertical lines calculated as \( y = \pm \frac{b^2}{c} \), resulting in locations at \( y = 3 \) and \( y = -3 \).

Graphing Ellipses

Graphing an ellipse requires plotting it correctly on the coordinate plane. Begin at the center, which for standard ellipses is \( (0,0) \).

• For the exercise's vertical ellipse:
  • Plot vertices using values \( (0, \pm \sqrt{3}) \) and \( (\pm \sqrt{2}, 0) \).
  • Mark foci at points \( (0,1) \) and \( (0,-1) \).
  • Draw directrices as horizontal lines at \( y = 3 \) and \( y = -3 \).
• Ellipses are symmetric about their axes, so carefully plot ensuring symmetry for accuracy.

These steps help visualize how the ellipse exists in space relative to its mathematical properties, displaying the characteristic elongated shape along the major axis.