Problem 1 - Thomas Calculus

Question

In Exercises \(1-8,\) find the eccentricity of the ellipse. Then find andgraph the ellipse's foci and directrices. $$16 x^{2}+25 y^{2}=400$$

Step-by-Step Solution

Answer

Eccentricity is \(\frac{3}{5}\). Foci are \((3,0)\) and \((-3,0)\); directrices are \(x=\frac{25}{3}\) and \(x=-\frac{25}{3}\).

Step 1: Express the Equation in Standard Form

Divide every term of the given equation \(16x^2 + 25y^2 = 400\) by 400 to get the standard form of an ellipse equation: \(\frac{x^2}{25} + \frac{y^2}{16} = 1\). This tells us that \(a^2 = 25\) and \(b^2 = 16\).

Step 2: Identify the Major and Minor Axes

Since \(a^2 = 25 > b^2 = 16\), the major axis is along the x-axis. This means the ellipse is horizontal with \(a = 5\) and \(b = 4\).

Step 3: Calculate the Eccentricity

The eccentricity \(e\) of an ellipse is given by \(e = \frac{c}{a}\), where \(c^2 = a^2 - b^2\). Thus, \(c^2 = 25 - 16 = 9\), so \(c = 3\). Therefore, \(e = \frac{3}{5}\).

Step 4: Determine the Foci

The foci of the ellipse are located at \((c, 0)\) and \((-c, 0)\) due to the horizontal major axis. Therefore, the foci are at \((3, 0)\) and \((-3, 0)\).

Step 5: Find the Equations of the Directrices

The directrices are vertical lines located at \(x = \pm \frac{a}{e}\). Substitute \(a = 5\) and \(e = \frac{3}{5}\) to find the directrices are at \(x = \pm \frac{5}{\frac{3}{5}} = \pm \frac{25}{3}\).

Step 6: Graph the Ellipse, Foci, and Directrices

To graph, draw the ellipse with center at the origin, horizontal stretching to \(\pm 5\) on the x-axis and vertical stretching to \(\pm 4\) on the y-axis. Mark the foci at \((3, 0)\) and \((-3, 0)\). Draw vertical lines at \(x = \frac{25}{3}\) and \(x = -\frac{25}{3}\) for the directrices.

Key Concepts

Standard Form of Ellipse

The standard form of an ellipse is a way to represent its equation in a clear and concise format. It helps us understand the basic parameters of the ellipse, like its dimensions and orientation. An ellipse's equation is given by \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where:

• \(a^2\) and \(b^2\) are the denominators under the square terms, representing the squared lengths of the ellipse's semi-major and semi-minor axes.
• \(a > b\) for a horizontal orientation, or \(b > a\) for a vertical orientation.

In the given problem, dividing each term by 400 transforms the initial equation \(16x^2 + 25y^2 = 400\) into the standard form \(\frac{x^2}{25} + \frac{y^2}{16} = 1\). This immediately tells us that the ellipse is centered at the origin with semi-major axis along the x-axis. Understanding this standard form is crucial for finding other properties, like axes lengths, foci, and directrices.

Major and Minor Axes

The major and minor axes are the longest and shortest diameters of the ellipse, respectively. They play a crucial role in defining the ellipse's shape and size.

• The **major axis** is associated with the larger denominator in the standard form equation, stretching the longest across the ellipse.
• The **minor axis** aligns with the smaller denominator.

From the equation \(\frac{x^2}{25} + \frac{y^2}{16} = 1\), we see that:

• \(a^2 = 25\) gives us the semi-major axis \(a = 5\).
• \(b^2 = 16\) gives us the semi-minor axis \(b = 4\).

Since \(a > b\), the longer axis is horizontal along the x-axis. Knowing the major and minor axes helps us draw the ellipse and provides important information about its geometric layout. It is also essential for locating the foci.

Foci and Directrices

The foci of an ellipse are two fixed points inside the ellipse used in its construction and defining its shape. The directrices are lines that are also part of this definition.**Foci**:

• The foci are located on the major axis, equidistant from the center.
• To find the foci, we use \(c^2 = a^2 - b^2\). Here \(c = 3\), so the foci are at \((3, 0)\) and \((-3, 0)\).

Foci help measure how 'stretched' an ellipse is. This stretch is quantified as eccentricity \(e\), calculated as \(e = \frac{c}{a} = \frac{3}{5}\), representing the ellipse's elongation.**Directrices**:

• These are vertical lines outside the ellipse that work with the eccentricity to define the ellipse.
• The directrices for this ellipse are given by \(x = \pm \frac{a}{e}\). Substituting known values, we find them at \(x = \pm \frac{25}{3}\).

Understanding the positioning of the foci and the lines of the directrices provides comprehensive insight into the geometric and algebraic nature of ellipses.