Problem 29 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the center, vertices, foci, and eccentricity of the ellipse, and sketch its graph. Use a graphing utility to verify your graph. $$\frac{x^{2}}{64}+\frac{y^{2}}{9}=1$$

Step-by-Step Solution

Answer

The center of the ellipse is at the origin (0,0). The vertices of the ellipse are at points (±8, 0), and the co-vertices are at points (0, ±3). The foci are located at points \((\pm \sqrt{55}, 0)\). The eccentricity of the ellipse is \(\frac{\sqrt{55}}{8}\).

Step 1: Identify the major axis and minor axis

The bigger denominator denotes the major axis and the smaller one denotes the minor axis. In the ellipse, \(\frac{x^{2}}{64}+\frac{y^{2}}{9}=1\), 'a' is the square root of 64 and 'b' is the square root of 9. Here, a=8 and b=3. Thus, the major axis is along the 'x' axis and minor axis is along the 'y' axis.

Step 2: Determine the center of the ellipse

For the given ellipse, the equation is in standard form with x and y not having any offsets, thus the center of ellipse is at origin, (0,0).

Step 3: Calculate the vertices and co-vertices of the ellipse

As the semi-major axis is on x-axis, vertices will be at (±a, 0), thus they are at points (±8, 0). Similarly, co-vertices will be at points (0, ±b) = (0,±3).

Step 4: Find the foci of the ellipse

The foci of the ellipse are at (±c,0) where c is given by \(c=\sqrt{a^{2}-b^{2}}\). Thus, \(c=\sqrt{64-9}=\sqrt{55}\). Therefore, foci are at points \((\pm \sqrt{55}, 0)\).

Step 5: Compute the eccentricity of the ellipse

Eccentricity, \(e=\frac{c}{a}=\frac{\sqrt{55}}{8}\)

Step 6: Sketch the graph of the ellipse

First, plot the center of the ellipse at the origin. Next, plot the vertices and co-vertices of the ellipse. Then, denote the foci on the diagram. Lastly, sketch the ellipse as a closed curve that passes through these points. Have a graphing utility to verify the graph.

Key Concepts

Center of an Ellipse

In the standard form of an ellipse equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]the center is found at the point where both x and y values are offset by zero, which is often given as (h, k). This means the center of the ellipse is located at (h, k). For the exercise example, the equation \[ \frac{x^2}{64} + \frac{y^2}{9} = 1 \]does not have any shifting parameters (h, k), which indicates the center is at the origin, (0, 0).

This center acts as the midpoint and measuring reference point for angles and distances within the ellipse. It's the starting point for determining the positions of the vertices and foci.

Vertices of an Ellipse

Vertices on an ellipse represent its widest and most narrow points, located along the major and minor axes. In our ellipse equation \[ \frac{x^2}{64} + \frac{y^2}{9} = 1 \],the major axis lies along the x-axis since 64 is larger than 9. The vertices are calculated by identifying the value 'a' in the denominator of the major axis term (here, a = 8).

The vertices are at coordinates \[(\pm a, 0) = (\pm 8, 0)\]. This shows that they extend 8 units left and right from the center along the x-axis.

For the co-vertices, since the minor axis lies along the y-axis, with 'b' identified as the square root of 9 (b = 3), these are located at \[(0, \pm b) = (0, \pm 3)\].This clarification of vertices helps in understanding the elliptical shape.

Foci of an Ellipse

The foci are special points inside the ellipse that help define its shape. To find the foci, we calculate 'c' using the formula \[c = \sqrt{a^2 - b^2}\]. In this example problem, 'a' is 8 and 'b' is 3, so\[c = \sqrt{64 - 9} = \sqrt{55}.\]

The foci are positioned along the major axis, at points given by \[(\pm c, 0) = (\pm \sqrt{55}, 0)\]. These points are inside the ellipse on the x-axis, confirming where the ellipse is stretched out.

Both foci are crucial for forming the elliptical shape, as every point on an ellipse has the same total distance to each focus.

Eccentricity of an Ellipse

Eccentricity is a measure of how "stretched" or "circular" an ellipse is. It's calculated using the formula \[e = \frac{c}{a}\],where 'c' is the distance from the center to a focus, and 'a' is the length of the semi-major axis. In the given equation, we calculated \[c = \sqrt{55}\] and 'a' is 8, leading to an eccentricity of \[e = \frac{\sqrt{55}}{8}.\]

Eccentricity values range from 0 to 1:

• A value closer to 0 indicates the ellipse is more circular.
• A value closer to 1 signifies a more elongated shape.

Eccentricity helps in understanding the overall proportionality of the ellipse, showing how much it deviates from being a perfect circle.