Problem 13 - Thomas Calculus

Question

Exercises \(13-16\) give foci and corresponding directrices of ellipses centered at the origin of the \(x y\) -plane. In each case, use the dimensions in Figure 10.19 to find the eccentricity of the ellipse. Then find the ellipse's standard-form equation. $$ \begin{array}{l}{\text { Focus: }(\sqrt{5}, 0)} \\ {\text { Directrix: } \quad x=\frac{9}{\sqrt{5}}}\end{array} $$

Step-by-Step Solution

Answer

Eccentricity is \(\frac{5}{9}\); equation is \(\frac{x^2}{\frac{81}{5}} + \frac{y^2}{16} = 1\)."

Step 1: Understand Definitions

In an ellipse, the distance from the center to a focus is denoted as \(c\), and the distance from the center to the directrix is \(d\). The eccentricity \(e\) of the ellipse is defined as \(e = \frac{c}{a}\), where \(a\) is the semi-major axis.

Step 2: Identify Given Values

From the focus \((\sqrt{5}, 0)\), we know that \(c = \sqrt{5}\). From the directrix \(x = \frac{9}{\sqrt{5}}\), we set \(d = \frac{9}{\sqrt{5}}\).

Step 3: Use Eccentricity Formula

The formula for eccentricity in relation to directrix is \(e = \frac{c}{d}\), which simplifies to \(e = \frac{\sqrt{5}}{\frac{9}{\sqrt{5}}} = \frac{5}{9}\).

Step 4: Solve for \(a\) Using Eccentricity

From \(e = \frac{c}{a}\), we have \(\frac{5}{9} = \frac{\sqrt{5}}{a}\). Solving for \(a\), we multiply both sides by \(a\) and then solve for \(a\), yielding \(a = \frac{9\sqrt{5}}{5}\).

Step 5: Calculate \(b\) Using Relationship with \(c, a\)

Using the relationship \(b^2 = a^2 - c^2\), substitute \(a = \frac{9\sqrt{5}}{5}\) and \(c = \sqrt{5}\) into the equation: \(b^2 = \left(\frac{9\sqrt{5}}{5}\right)^2 - (\sqrt{5})^2 = \frac{81 \cdot 5}{25} - 5 = \frac{400}{25} = 16\). So, \(b = \sqrt{16} = 4\).

Step 6: Write Standard Form Equation

The standard form equation for an ellipse centered at the origin with semi-major axis \(a\) and semi-minor axis \(b\) is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Substituting \(a = \frac{9\sqrt{5}}{5}\) and \(b = 4\), we get the equation: \(\frac{x^2}{\left(\frac{9\sqrt{5}}{5}\right)^2} + \frac{y^2}{16} = 1\), which simplifies to \(\frac{x^2}{\frac{81}{5}} + \frac{y^2}{16} = 1\).

Key Concepts

Eccentricity

The eccentricity of an ellipse is a crucial concept that measures its deviation from being circular. In simpler terms, eccentricity, denoted by the symbol \(e\), tells us how "stretched" the ellipse is.
- When \(e = 0\), the figure is a perfect circle. - As \(e\) approaches 1, the ellipse becomes more elongated.
Eccentricity is always a value between 0 and 1 for ellipses. It's defined as the ratio of the distance from the center of the ellipse to one of its foci (denoted as \(c\)) over the length of the semi-major axis (denoted as \(a\)). This is given by the formula:\[e = \frac{c}{a}\]In the provided exercise, the eccentricity is calculated using the relationship of the focus and directrix, yielding \(e = \frac{5}{9}\). This tells us that the ellipse is somewhat elongated, but not overly so.

Foci of an Ellipse

The foci (plural of focus) are two fixed points within an ellipse that help define its shape. Imagine that each focus has an invisible string attached to any point on the ellipse, keeping it tied to these fixed points.
- The sum of the distances from any point on the ellipse to the two foci is constant.The coordinates of the foci depend on the orientation of the ellipse. For an ellipse centered at the origin, if the major axis is along the x-axis, the foci are at \((\pm c, 0)\). If the major axis is along the y-axis, the foci would be at \((0, \pm c)\).
In the example provided, one of the foci is at \((\sqrt{5}, 0)\), indicating that the major axis is horizontal. The second focus is symmetrically placed at \((-\sqrt{5}, 0)\). The distance \(c = \sqrt{5}\) helps us in determining the ellipse's eccentricity and overall shape.

Ellipse Equation

The standard equation for an ellipse centered at the origin reveals its dimensions and orientation in the coordinate plane. For an ellipse, this equation is\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. These axes determine how "wide" and "tall" the ellipse is. If \(a > b\), the ellipse stretches horizontally, and if \(b > a\), it stretches vertically.
In this exercise, the values found for \(a\) and \(b\) enable us to form the ellipse's equation. Here, with \(a = \frac{9\sqrt{5}}{5}\) and \(b = 4\), the equation becomes\[\frac{x^2}{\left(\frac{81}{5}\right)} + \frac{y^2}{16} = 1\]This indicates the ellipse is wider horizontally, following the line of the semi-major axis since \(a > b\). The standard form equation is key in graphing ellipses and understanding their dimensions.

Directrix

The directrix of an ellipse is a line used in defining the ellipse’s eccentricity. It is one of the fixed guidelines that binds the points on the ellipse along with the focus.
- Directrix acts as a reference line in the calculation of the eccentricity.When dealing with the directrix, the eccentricity \(e\) can be found using the formula:\[e = \frac{c}{d}\]where \(d\) is the distance from the center to the directrix. The ellipse's directrix is positioned outside the ellipse on the side opposite to the foci; however, this may change based on its position.
From the example exercise, the directrix is given by \(x = \frac{9}{\sqrt{5}}\). Such a value confirms that the directrix helps determine the position of the elliptical shape in relation to its set point of origin. Understanding the role of a directrix not only aids in defining eccentricity but also influences how accurately one can sketch and comprehend ellipses.