Problem 17 - 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. $$ x^{2}+4 y^{2}=1 $$

Step-by-Step Solution

Answer

Vertices: (±1, 0), Foci: (±√3/2, 0), Eccentricity: √3/2, Major axis: 2, Minor axis: 1.

Step 1: Identify the standard form of the ellipse

The standard form of an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) for an ellipse centered at the origin. Given the equation \( x^2 + 4y^2 = 1 \), we can rewrite it as \( \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1 \). This helps us identify \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \).

Step 2: Determine the semi-major and semi-minor axes lengths

In the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), if \( a^2 > b^2 \), then \( a \) is the semi-major axis and \( b \) is the semi-minor axis. For the equation \( \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1 \), we have \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \), so \( a = 1 \) and \( b = \frac{1}{2} \). This indicates the semi-major axis is \( a = 1 \) and the semi-minor axis is \( b = \frac{1}{2} \).

Step 3: Calculate the vertices of the ellipse

The vertices of the ellipse will be along the x-axis since \( a > b \) and it's centered at the origin. The vertices are located at \((\pm a, 0)\). With \( a = 1 \), the vertices are \((1, 0)\) and \((-1, 0)\).

Step 4: Calculate the foci of the ellipse

The foci are determined using the formula \( c^2 = a^2 - b^2 \). Here, \( c^2 = 1 - \frac{1}{4} = \frac{3}{4} \) which gives \( c = \frac{\sqrt{3}}{2} \). Thus, the foci are \((\pm \frac{\sqrt{3}}{2}, 0)\).

Step 5: Find the eccentricity of the ellipse

The eccentricity \( e \) is given by \( e = \frac{c}{a} \). Substitute \( c = \frac{\sqrt{3}}{2} \) and \( a = 1 \) to get \( e = \frac{\sqrt{3}}{2} \).

Step 6: Determine the lengths of the major and minor axes

The length of the major axis is \( 2a \) and the minor axis is \( 2b \). Hence, the major axis length is \( 2 \times 1 = 2 \) and the minor axis length is \( 2 \times \frac{1}{2} = 1 \).

Step 7: Sketch the ellipse

Draw the x-axis and y-axis on a graph. Plot the vertices at \( (1,0) \) and \( (-1,0) \). Plot the endpoints of the minor axis at \( (0, \frac{1}{2}) \) and \( (0, -\frac{1}{2}) \). Sketch an ellipse that passes through these points, stretched along the x-axis.

Key Concepts

Vertices

Vertices are specific points on an ellipse that indicate the farthest reaches of the ellipse along its major axis. These are important because they help define the shape and size of the ellipse.
When an ellipse is positioned with its center at the origin of a coordinate system, its vertices can be easily found. For the ellipse with the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the vertices are located along the major axis.

• If the major axis is horizontal: Vertices are at \((\pm a, 0)\).
• If the major axis is vertical: Vertices are at \((0, \pm a)\).

In our example, the given equation \( x^2 + 4y^2 = 1 \) can be transformed into the standard form \( \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1 \). Thus, \( a^2 = 1 \), which means \( a = 1 \). Because \( a > b \), the major axis is horizontal, and the vertices are at \((1, 0)\) and \((-1, 0)\).

Foci

The foci (plural of focus) of an ellipse are two crucial points inside the ellipse through which the sum of the distances to any point on the ellipse is constant. These points help to define the ellipse’s elongated shape.
To find the foci, we use the relationship \( c^2 = a^2 - b^2 \), where \( c \) is the distance from the center to each focus.
Let's find the foci for the ellipse given by \( x^2 + 4y^2 = 1 \). With \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \), we calculate:

• \( c^2 = 1 - \frac{1}{4} = \frac{3}{4} \)
• \( c = \frac{\sqrt{3}}{2} \)

Since the major axis is horizontal, the foci are positioned along the x-axis at \((\pm \frac{\sqrt{3}}{2}, 0)\).
This means you locate one focus to the right of the center and one to the left, equal in distance.

Eccentricity

Eccentricity is a measure that describes how "stretched" an ellipse is around its foci. It gives an indication of the deviation of the ellipse from being a perfect circle.
The formula for eccentricity \( e \) of an ellipse is \( e = \frac{c}{a} \). Here, \( c \) is the distance from the center to a focus, and \( a \) is the semi-major axis.

• When the eccentricity \( e = 0 \): The ellipse is a circle.
• When \( 0 < e < 1 \): The ellipse is more stretched out.

For the ellipse \( x^2 + 4y^2 = 1 \), we have already found \( c = \frac{\sqrt{3}}{2} \) and \( a = 1 \). Thus, the eccentricity is:\[ e = \frac{\sqrt{3}}{2} \]This eccentricity tells us that our ellipse is elongated along the horizontal direction but is not extremely stretched. It lies between a circle and a parabola, showing its unique shape.

Major and Minor Axes

The major and minor axes are lines through the center of an ellipse. They correspond to its longest and shortest diameters, respectively.
In the context of the ellipse formula \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):

• The **major axis** has a length of \( 2a \). It is the longest diameter from vertex to vertex.
• The **minor axis** spans \( 2b \). It's the shortest distance across the ellipse.

For our example with the equation \( x^2 + 4y^2 = 1 \), we determined that \( a = 1 \) and \( b = \frac{1}{2} \). Hence, the lengths are:

• Major axis length: \( 2 \times 1 = 2 \)
• Minor axis length: \( 2 \times \frac{1}{2} = 1 \)

This means the ellipse is twice as long in its major axis direction, giving it a more "stretched" appearance horizontally.