Problem 5 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Eccentricity is \(\frac{1}{\sqrt{3}}\), foci: \((0, \pm 1)\), directrices: \(y = \pm 3\).

Step 1: Write the Equation in Standard Form

The given equation of the ellipse is \(3x^2 + 2y^2 = 6\). To write it in standard form, divide every term by 6: \[\frac{3x^2}{6} + \frac{2y^2}{6} = 1\]. We get \(\frac{x^2}{2} + \frac{y^2}{3} = 1\).

Step 2: Identify the Necessary Parameters

The standard form of an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a^2 = 2\) and \(b^2 = 3\). Since \(b^2 > a^2\), it indicates a vertical ellipse. Therefore, \(a = \sqrt{2}\), and \(b = \sqrt{3}\).

Step 3: Calculate Eccentricity

For an ellipse, the eccentricity \(e\) is given by \(e = \sqrt{1 - \frac{a^2}{b^2}}\). Substitute \(a^2 = 2\) and \(b^2 = 3\), so: \[e = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\].

Step 4: Find the Foci

The foci \((0,\pm c)\) of an ellipse are found using \(c = be\). Since \(b = \sqrt{3}\) and \(e = \frac{1}{\sqrt{3}}\), we get \(c = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1\). Thus, the foci are \((0, \pm 1)\).

Step 5: Find the Directrices

The equations of the directrices for an ellipse are given by \(y = \pm \frac{b^2}{c}\). Using \(b^2 = 3\) and \(c = 1\), the directrices are \(y = \pm 3\).

Key Concepts

Ellipse Standard Form

In mathematics, an ellipse can often be described by a specific type of equation, known as the ellipse's standard form. This standard form helps in identifying the shape and orientation of the ellipse. It is generally written as \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] Where:

• \(a\) and \(b\) are constants that determine the length of the semi-major and semi-minor axes.
• The ellipse is horizontal if \(a^2 > b^2\) and vertical if \(b^2 > a^2\).

For example, given the equation \(3x^2 + 2y^2 = 6\), by dividing all terms by 6, we transform it to: \[\frac{x^2}{2} + \frac{y^2}{3} = 1\]. This indicates that \(a^2 = 2\) and \(b^2 = 3\), implying a vertical ellipse since \(b^2 > a^2\). Knowing how to properly write the equation in standard form makes it much easier to further analyze and graph the ellipse.

Foci of an Ellipse

The foci of an ellipse are two special points located along the major axis, and they play a critical role in the geometric definition of an ellipse. The total distance from any point on the ellipse to each focus is constant. For ellipses, the foci are deduced by determining the value of \(c\), calculated using: \[c = b \times e \] where \(e\) is the eccentricity of the ellipse. The eccentricity \(e\) is obtained through \[e = \sqrt{1 - \frac{a^2}{b^2}}\]. In our example, substitute \(a^2 = 2\) and \(b^2 = 3\) to get \[e = \sqrt{\frac{1}{3}}\]. Thus, \(c = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1\). This indicates that the foci are located at the points \((0, \pm 1)\) on the ellipse. Understanding the position and significance of the foci enables deeper geometric insights and supports tasks like plotting the ellipse."},{"concept_headline":"Directrices of an Ellipse","text":"Directrices are mathematical lines associated with ellipses, extending the concept of directrix from parabolas. Each ellipse is connected to two directrices, which helps in understanding its shape and properties in relation to its eccentricity. The directrices of a vertical ellipse are vertical lines described by the equation: \(y = \pm \frac{b^2}{c}\). In our case, with \(b^2 = 3\) and \(c = 1\), we find the directrices as \(y = \pm 3\). Thus, the directrices are lines \(y = 3\) and \(y = -3\).By knowing the position of the directrices, students can ensure they accurately graph the shape of the ellipse based on its core geometric traits and relationships with its foci."}]}}]}]}]}}"]}]} ]}]}]}]}]}]}]}]}]} Node backend worker failed with exit code 247 on message {'user_pull_request_id': 'dcea5147-110d-4cc5-86fc-074c3be24080', 'respond_to': {'thread_id': 'XXX-516'}