Problem 6 - Algebra and Trigonometry

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$ \frac{x^{2}}{16}+\frac{y^{2}}{25}=1 $$

Step-by-Step Solution

Answer

Vertices: (0, 5) and (0, -5); Foci: (0, 3) and (0, -3); Eccentricity: 0.6; Major Axis: 10; Minor Axis: 8.

Step 1: Identify the form of the ellipse

The given ellipse equation is \( \frac{x^2}{16} + \frac{y^2}{25} = 1 \). This is in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( b^2 = 25 \) and \( a^2 = 16 \). Since \( b^2 > a^2 \), this is a vertical ellipse.

Step 2: Determine lengths of axes

For ellipses, the lengths of the axes are determined by \( 2a \) and \( 2b \). Here, \( a = \sqrt{16} = 4 \) and \( b = \sqrt{25} = 5 \). Therefore, the lengths are: Major Axis = \( 2b = 10 \), Minor Axis = \( 2a = 8 \).

Step 3: Identify vertices

For a vertical ellipse, vertices are located at \( (0, \pm b) \). Here, \( b = 5 \), so the vertices are \( (0, 5) \) and \( (0, -5) \).

Step 4: Calculate eccentricity

The eccentricity \( e \) of an ellipse is given by \( e = \frac{\sqrt{b^2-a^2}}{b} \). Substituting \( a^2 = 16 \) and \( b^2 = 25 \), we get \( e = \frac{\sqrt{25-16}}{5} = \frac{3}{5} \).

Step 5: Find foci

The foci of a vertical ellipse are located at \( (0, \pm c) \) where \( c = \sqrt{b^2-a^2} \). Here, \( c = \sqrt{25-16} = 3 \). Therefore, the foci are \( (0, 3) \) and \( (0, -3) \).

Step 6: Sketch the Graph

To sketch the graph, draw the ellipse centered at the origin (0,0), mark the vertices on the y-axis at (0, 5) and (0, -5), and the foci at (0, 3) and (0, -3). The endpoints of the minor axis are on the x-axis at (4, 0) and (-4, 0). This setup highlights the vertical nature of the ellipse.

Key Concepts

Vertices

Vertices are the key points that help to define the shape and orientation of an ellipse. In simplest terms, vertices are the points where the ellipse is at its widest. In a vertical ellipse, they are located along the vertical axis.
For the ellipse defined by the equation \( \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 \), we have a vertical ellipse. This means the vertices are situated on the y-axis because the term with \( y \) has the larger denominator.
To find the vertices of this specific ellipse, look at the values of \( b \). Here, \( b = 5 \) (since \( b^2 = 25 \)). The vertices, therefore, are at \( (0, 5) \) and \( (0, -5) \).
These are the highest and lowest points that the ellipse will reach along the y-axis.

Foci

The foci (plural of focus) are special points found within an ellipse. They are essential in defining the elliptical shape. The sum of the distances from any point on the ellipse to each of the foci is constant.
In our vertical ellipse \( \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 \), the foci are located along the major axis, which is the y-axis in this case.
To find the distance of the foci from the center, we use the formula \( c = \sqrt{b^2 - a^2} \). Substituting the known values, with \( b^2 = 25 \) and \( a^2 = 16 \), we find \( c = 3 \).
Thus, the foci are located at \( (0, 3) \) and \( (0, -3) \), positioned symmetrically along the y-axis, inside the ellipse.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. It is a number between 0 and 1, where 0 indicates a perfect circle and values closer to 1 indicate a more elongated shape.
For an ellipse, the eccentricity \( e \) is calculated as \( e = \frac{\sqrt{b^2 - a^2}}{b} \).
In our ellipse \( \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 \), we substitute \( b^2 = 25 \) and \( a^2 = 16 \) to find \( e = \frac{\sqrt{9}}{5} = \frac{3}{5} \).
This value of \( \frac{3}{5} \) indicates that the ellipse is moderately elongated and not too circular. Lower eccentricity would mean it is more circular, while a higher value would indicate it is more stretched out.

Major Axis

The major axis is the longest diameter of the ellipse. It stretches from one end of the ellipse to the other, passing through both vertices, and includes the foci.
In a vertical ellipse like \( \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 \), the major axis is aligned with the y-axis.
The length of the major axis is calculated using the formula \( 2b \). For this ellipse, since \( b = 5 \), the major axis measures \( 2 \times 5 = 10 \).
This axis is crucial because it provides the longest extent of the ellipse and helps us understand its orientation with respect to its center.

Minor Axis

The minor axis is the shortest diameter of the ellipse. Like the major axis, it passes through the center of the ellipse, but it is perpendicular to the major axis.
For the vertical ellipse given by \( \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 \), the minor axis lies along the x-axis.
The length of the minor axis is calculated using \( 2a \). Here, \( a = 4 \) (since \( a^2 = 16 \)), so the minor axis is \( 2 \times 4 = 8 \) in length.
This axis, being shorter, helps to highlight the elliptical nature of the figure by offering a contrast to the longer major axis. Together, these axes help determine the overall shape and appearance of the ellipse.