Problem 19 - Thomas Calculus

Question

The endpoints of the major and minor axes of an ellipse are \((1,1),\) \((3,4),(1,7),\) and \((-1,4) .\) Sketch the ellipse, give its equation in standard form, and find its foci, eccentricity, and directrices.

Step-by-Step Solution

Answer

An ellipse centered at (2, 2.5) with equation \(\frac{(x-2)^2}{\frac{13}{4}} + \frac{(y-2.5)^2}{\frac{9}{4}} = 1\), foci at (3, 2.5) and (1, 2.5), eccentricity \(\frac{2}{\sqrt{13}}\), and directrices at \(x = \frac{21}{4}\) and \(x = \frac{-5}{4}\).

Step 1: Identify the Axes

First, we identify the major and minor axes of the ellipse. Given points \((1,7)\) and \((-1,4)\), the distance between them (2 units) is less than the distance between \((1,1)\) and \((3,4)\), which is approximately 3.6 units. This means \((1,1)\) to \((3,4)\) is the major axis.

Step 2: Calculate the Center

The center of the ellipse \((h,k)\) is the midpoint of the endpoints of the major axis \((1,1)\) and \((3,4)\). Use the midpoint formula \((h,k) = \left( \frac{1+3}{2}, \frac{1+4}{2} \right) = \left(2, 2.5\right)\).

Step 3: Determine Semi-Major and Semi-Minor Axes

The semi-major axis \(a\) is half the length of the major axis. Using endpoints \((1,1)\) and \((3,4)\), the distance is \(\sqrt{(3-1)^2 + (4-1)^2} = \sqrt{13}\), so \(a = \frac{\sqrt{13}}{2}\). For the semi-minor axis \((1,7)\) and \((-1,4)\), the distance is 3, so \(b = \frac{3}{2}\).

Step 4: Write the Ellipse Equation

In standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we substitute\(h=2, k=2.5, a=\frac{\sqrt{13}}{2}, b=\frac{3}{2}\). The equation becomes \(\frac{(x-2)^2}{\left(\frac{{\sqrt{13}}}{2}\right)^2} + \frac{(y-2.5)^2}{\left(\frac{3}{2}\right)^2} = 1\),which simplifies to \(\frac{(x-2)^2}{\frac{13}{4}} + \frac{(y-2.5)^2}{\frac{9}{4}} = 1\).

Step 5: Calculate Foci

The distance to the foci \(c\) is found using \(c^2 = a^2 - b^2\). Thus, \(c^2 = \frac{13}{4} - \frac{9}{4} = 1\) giving \(c = 1\). The foci are at \((2 \pm c, 2.5) = (3, 2.5)\) and \((1, 2.5)\).

Step 6: Determine Eccentricity

The eccentricity of an ellipse is given by \(\frac{c}{a} = \frac{1}{\frac{\sqrt{13}}{2}} = \frac{2}{\sqrt{13}}\).

Step 7: Find the Directrices

The directrices are located at \(x = h \pm \frac{a^2}{c}\).Thus: \(x = 2 \pm \frac{\frac{13}{4}}{1}\). This simplifies to \(x = 2 \pm \frac{13}{4}\), resulting in \(x = \frac{21}{4}\) and \(x = \frac{-5}{4}\).

Step 8: Sketch the Ellipse

Using the calculated center \((2, 2.5)\), the semi-major axis \(a = \frac{\sqrt{13}}{2}\), and semi-minor axis \(b = \frac{3}{2}\), sketch the ellipse. The figure will be elongated along the line connecting \((1,1)\) to \((3,4)\) with the calculated foci and directrices.

Key Concepts

Major Axis

In the context of an ellipse, the major axis is its longest diameter. It is the line that runs through the center and both foci of the ellipse, extending to the widest points of the shape. In the given problem, the points

• \((1,1)\)
• \((3,4)\)

represent the endpoints of the major axis. The calculation of its length helps us understand the overall size of the ellipse. We find that the length of the major axis is approximately 3.6 units, which is greater than the length of the minor axis. Identifying this axis is crucial because it helps us determine related parameters such as the semi-major axis and the orientation of the ellipse.

Minor Axis

The minor axis is the shortest diameter of the ellipse, perpendicular to the major axis. It also runs through the center but does not extend to the foci. Instead, it stretches from one side of the ellipse to the other at its narrowest point. In the exercise scenario,

• \((1,7)\)
• \((-1,4)\)

are the endpoints of the minor axis. The length of the minor axis is 3, which is shorter than the major axis. Just like with the major axis, learning the minor axis is fundamental in determining the complete geometry of the ellipse, allowing us to decide on the semi-minor axis essential for drafting the ellipse's equation.

Foci

An essential attribute of an ellipse, the foci (plural of focus) are two distinct points located symmetrically along the major axis. They function as reference points that determine the shape of the ellipse.

• For our ellipse, once the center and dimensions of the axes are identified, we compute the distance to the foci, which is given by \(c = \sqrt{a^2 - b^2}\).
• In the specific problem, the value of \(c\) is 1, leading us to pinpoint the foci at points \((3, 2.5)\) and \((1, 2.5)\).

These are crucial when visualizing the ellipse because they directly affect how much the ellipse "spreads out" from its center along the major axis.

Eccentricity

Eccentricity is a numerical measure that describes how much an ellipse deviates from being a circle. It is calculated by the ratio\[e = \frac{c}{a}\]where \(c\) is the distance from the center to a focus point, and \(a\) is the semi-major axis.

• In this case, we find \(e = \frac{2}{\sqrt{13}}\), indicating the ellipse is not a perfect circle since a perfect circle would have an eccentricity of zero.

Understanding the eccentricity gives insights into the shape's nature — the closer the eccentricity is to 1, the more elongated the ellipse. For students, calculating eccentricity can offer a clear depiction of the ellipse's shape and its distinguishing features.

Directrices

The directrices are lines associated with an ellipse that provide another way to understand its geometry. They are positioned at a distance determined by \[x = h \pm \frac{a^2}{c}\]where \(h\) is the x-coordinate of the center, \(a\) is the semi-major axis length, and \(c\) is the distance to the foci.

• In the exercise, the directrices are found at \(x = \frac{21}{4}\) and \(x = \frac{-5}{4}\).

These lines help describe how each point on the ellipse relates geometrically, offering a foundational understanding of the ellipse's construction for both advanced mathematical applications and simpler conceptual learning.

Ellipse Equation

The equation of an ellipse is a fundamental representation that describes all points composing the ellipse. It is expressed in its standard form: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where

• \((h,k)\) represents the center coordinates
• \(a\) is the semi-major axis length, and
• \(b\) is the semi-minor axis length.

For the given exercise, inserting the calculated values leads to the formula \[\frac{(x-2)^2}{\frac{13}{4}} + \frac{(y-2.5)^2}{\frac{9}{4}} = 1\]This mathematical expression is vital for drawing accurate conclusions regarding the ellipse's shape, location, and other properties. Being comfortable working with this equation enhances analytical skills vital in more advanced studies.