Problem 2 - Thomas Calculus

Question

In Exercises \(1-8,\) find the eccentricity of the ellipse. Then find andgraph the ellipse's foci and directrices. $$7 x^{2}+16 y^{2}=112$$

Step-by-Step Solution

Answer

Eccentricity is \(\frac{3}{4}\). Foci are at \((3, 0)\) and \((-3, 0)\). Directrices are \(x = \frac{16}{3}\) and \(x = -\frac{16}{3}\).

Step 1: Simplify the Equation

Start by simplifying the given equation for the ellipse, \(7x^2 + 16y^2 = 112\). Divide every term by 112 to convert it into standard form. This gives: \[ \frac{x^2}{16} + \frac{y^2}{7} = 1 \] which is an equation of an ellipse.

Step 2: Identify the Semi-major Axes

The equation \( \frac{x^2}{16} + \frac{y^2}{7} = 1 \) is in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a^2 = 16\) and \(b^2 = 7\). Here, \(a = 4\) and \(b = \sqrt{7}\). Since \(a > b\), this is a horizontal ellipse.

Step 3: Calculate Eccentricity

For an ellipse, the eccentricity \(e\) is given by \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Plugging in the values, we get: \[ e = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]. Thus, the eccentricity is \(\frac{3}{4}\).

Step 4: Find the Foci

The coordinates of the foci for a horizontal ellipse are \((\pm c, 0)\) where \(c = ae\). Here, \(c = 4 \cdot \frac{3}{4} = 3\). Thus, the foci are at \((3, 0)\) and \((-3, 0)\).

Step 5: Determine the Directrices

The directrices for a horizontal ellipse can be found by using \(x = \pm \frac{a}{e}\). Plug in the values: \[ x = \pm \frac{4}{\frac{3}{4}} = \pm \frac{16}{3} \]. Thus, the equations of the directrices are \(x = \frac{16}{3}\) and \(x = -\frac{16}{3}\).

Step 6: Graph the Ellipse with Foci and Directrices

Graph the ellipse centered at the origin with a semi-major axis of length 4 along the x-axis and a semi-minor axis of length \(\sqrt{7}\) along the y-axis. Place the foci at \((3, 0)\) and \((-3, 0)\), and draw the vertical lines at \(x = \frac{16}{3}\) and \(x = -\frac{16}{3}\) for the directrices.

Key Concepts

Ellipse Equation

An ellipse is a fascinating shape, somewhat like a stretched circle. The equation of an ellipse in its standard form is either \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] or \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \] depending on its orientation. Here, "a" and "b" are the lengths of the semi-major and semi-minor axes. This standard form equation helps in easily identifying these axes and determining the ellipse's properties.

• For horizontal ellipses, the first form is used with \(a > b\).
• For vertical ellipses, the second form is used with \(b > a\).

Starting with a general second-degree equation of the form \(Ax^2 + By^2 + Cx + Dy + E = 0\), we often need to simplify it to the standard form by completing the square and dividing through by constants to ensure everything equals 1.

Major and Minor Axes

In an ellipse, the major axis is the longest diameter and the minor axis is the shortest. The length of the semi-major axis is denoted by "a" and the length of the semi-minor axis by "b".

• The major axis spans from one end of the ellipse to the other, passing through the center.
• It determines the longest direction of the ellipse.
• The semi-major axis extends from the center to the ellipse's furthest boundary along this line.

Similarly, the minor axis is perpendicular to the major axis at the center. The semi-minor axis extends from the center to the boundary in the direction of the minor axis. In the given equation \( \frac{x^2}{16} + \frac{y^2}{7} = 1 \), since \(a = 4\) and \(b = \sqrt{7}\), the ellipse is horizontal because \(a > b\).

Foci of an Ellipse

The foci (plural of focus) of an ellipse are two special points situated along the major axis. The distances of any point on the ellipse to each focus sum up to a constant value, unique to that ellipse. Think of the foci as anchors that shape the ellipse.

• The distance from the center to each focus is denoted by "c".
• In a horizontal ellipse, the foci are located at \((\pm c, 0)\).
• The value of "c" is calculated using the formula \(c = \sqrt{a^2 - b^2}\).

For the given ellipse, we find \(c\) by taking \(a = 4\) and \(b = \sqrt{7}\), leading to \(c = 3\). Therefore, the foci are at \((3, 0)\) and \((-3, 0)\).

Directrices of an Ellipse

Directrices are lines that provide another way to define an ellipse. Every ellipse has two directrices, corresponding to its foci. These provide additional guidance in understanding the ellipse's eccentricity.

• The equation for the directrices of an ellipse is \(x = \pm \frac{a}{e}\) for horizontal ellipses.
• "e" is the eccentricity of the ellipse, given by \(e = \sqrt{1 - \frac{b^2}{a^2}}\).

For eccentricity \(e = \frac{3}{4}\) in the given problem, the directrices are calculated as \(x = \pm \frac{16}{3}\). These lines help describe the "stretch" of the ellipse, showing how far it is from being a perfect circle.