Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(\frac{(x-2)^{2}}{64}+\frac{(y-4)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The center is (2, 4); vertices are (10, 4) and (-6, 4); foci are at \((2\pm4\sqrt{3}, 4)\).

Step 1: Identify the Standard Form of the Ellipse

The equation given is \( \frac{(x-2)^{2}}{64} + \frac{(y-4)^{2}}{16} = 1 \). This is of the form \( \frac{(x-h)^{2}}{a^2} + \frac{(y-k)^{2}}{b^2} = 1 \) where the ellipse is oriented horizontally because \( a^2 > b^2 \). Here, \( h=2 \), \( k=4 \), \( a^2=64 \), and \( b^2=16 \).

Step 2: Determine the Center

The center of the ellipse, \((h, k)\), can be found directly from the equation. Substitute \( h = 2 \) and \( k = 4 \). Therefore, the center of the ellipse is \( (2, 4) \).

Step 3: Calculate the Vertices

For a horizontally oriented ellipse, the vertices are located \( a \) units away from the center along the x-axis. Since \( a^2 = 64 \), \( a = 8 \). Thus, the vertices are \( (2+8, 4) = (10, 4) \) and \( (2-8, 4) = (-6, 4) \).

Step 4: Calculate the Co-Vertices

The co-vertices are located \( b \) units away from the center along the y-axis, where \( b^2 = 16 \) giving \( b = 4 \). They are \( (2, 4+4)=(2, 8) \) and \( (2, 4-4)=(2, 0) \).

Step 5: Find the Foci

For ellipses, the foci are determined by \( c^2 = a^2 - b^2 \). Substituting the known values, we get \( c^2 = 64 - 16 = 48 \), so \( c = \sqrt{48} = 4\sqrt{3} \). The foci, located \( c \) units from the center along the x-axis, are at \( (2+4\sqrt{3}, 4) \) and \( (2-4\sqrt{3}, 4) \).

Step 6: Sketch the Ellipse

Plot the center at \( (2, 4) \). Mark the vertices at \( (10, 4) \) and \( (-6, 4) \). Locate the co-vertices at \( (2, 8) \) and \( (2, 0) \). Draw the ellipse shape passing through these points, and mark the foci \( (2+4\sqrt{3}, 4) \) and \( (2-4\sqrt{3}, 4) \) inside the ellipse.

Key Concepts

Conic Sections

Conic sections are a fascinating topic in mathematics, involving the intersection of a plane and a double-napped cone. These sections result in various shapes, primarily circles, ellipses, parabolas, and hyperbolas. Ellipses, which are the focus of our discussion here, have unique properties and appear in different real-world contexts, such as planetary orbits.

The conic sections we see depend on the angle intersecting the cone:

• A circle is formed when the intersecting plane is perpendicular to the cone's axis.
• An ellipse appears when the plane cuts through the cone at an angle, neither perpendicular nor parallel to the cone's axis.
• A parabola is seen when the intersecting plane is parallel to the cone's slant.
• A hyperbola is formed when the plane intersects both nappes.

Ellipses are known for their symmetrical shape, often described as an "oval," and have two focal points which are crucial in their formation and properties.

Ellipse Properties

Understanding the properties of ellipses helps in recognizing their structure and behavior. An ellipse is described by its semi-major and semi-minor axes, which define its height and width.

- The **center** of the ellipse is found at the point \( (h, k) \), which is directly extracted from the ellipse's standard form equation.- The **semi-major axis** (\