Problem 19 - 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. $$ \frac{1}{2} x^{2}+\frac{1}{8} y^{2}=\frac{1}{4} $$

Step-by-Step Solution

Answer

Vertices are at (0,±2); foci are at (0,±1.87); eccentricity is \(\sqrt{0.875}\).

Step 1: Rewrite the Equation in Standard Form

The given equation is \( \frac{1}{2} x^{2} + \frac{1}{8} y^{2} = \frac{1}{4} \). First, multiply through by 4 to clear the fractions: \( 2x^2 + \frac{1}{2}y^2 = 1 \). Next, divide every term by 1 to leave the right side as 1, giving us the standard form: \( \frac{x^2}{\frac{1}{2}} + \frac{y^2}{4} = 1 \). This can be rewritten as: \( \frac{x^2}{0.5} + \frac{y^2}{4} = 1 \).

Step 2: Identify the Major and Minor Axes

In the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the larger denominator indicates the major axis. Here, \( b^2 = 4 > a^2 = 0.5 \), so the ellipse has a vertical major axis. The lengths of the axes are determined by \( 2b \) and \( 2a \).

Step 3: Calculate Major and Minor Axis Lengths

Calculate the lengths by taking the square roots of the denominators: \( a = \sqrt{0.5} = \frac{1}{\sqrt{2}} \) or \( \frac{\sqrt{2}}{2} \), so the minor axis length is \( 2a = \sqrt{2} \). For \( b, b^2 = 4 \) so \( b = 2 \), making the major axis length \( 2b = 4 \).

Step 4: Find the Eccentricity

The eccentricity of an ellipse is calculated using \( e = \sqrt{1 - \frac{a^2}{b^2}} \). Here, \( a^2 = 0.5 \) and \( b^2 = 4 \), so \( e = \sqrt{1 - \frac{0.5}{4}} = \sqrt{1 - 0.125} = \sqrt{0.875} \).

Step 5: Determine the Vertices and Foci

Vertices lie along the major axis at \( (0, \pm b) = (0, \pm 2) \). The distance to the foci from the center is \( c = e \times b \). Compute \( c = \sqrt{0.875} \times 2 \approx 1.87 \). Thus, the foci are at \( (0, \pm 1.87) \).

Step 6: Sketch the Graph

Plot a vertical ellipse centered at \((0,0)\) with vertices at \((0, \pm 2)\) and minor endpoints at roughly \((\pm \sqrt{2}/2, 0)\). The foci at \((0, \pm 1.87)\) indicate the elongation along the vertical axis.

Key Concepts

Vertices of Ellipse

In an ellipse, the vertices are critical points located along the longest diameter, known as the major axis. These vertices mark the farthest extent of the ellipse from its center.
For the ellipse we are exploring, described by the equation \( \frac{x^2}{0.5} + \frac{y^2}{4} = 1 \), it has a major vertical axis.
This means our vertices are positioned vertically from the center, as the equation is symmetric about the y-axis.
The vertices are located at \((0, \pm b)\), where \(b\) is the square root of our larger denominator.
Therefore, since \(b = 2\), the vertices of this ellipse can be found at \((0, 2)\) and \((0, -2)\). These points are equidistant from the center, marking the peak of the ellipse along its major stretch.

Foci of Ellipse

The foci of an ellipse are internal points aligned along the major axis that help define its shape.
These points are instrumental because they provide a unique property of ellipses: the sum of the distances from any point on the ellipse to the two foci is constant.
The distance from the center of the ellipse to each focus is given by \(c = e \times b\), where \(e\) represents the eccentricity.
For this ellipse, we found that \(b = 2\) and the eccentricity \(e = \sqrt{0.875} = 0.935\).
Therefore, the distance \(c\) is approximately \(1.87\).
Thus, the foci are located at \((0, \pm 1.87)\), extending vertically from the center of the ellipse.

Eccentricity of Ellipse

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle.
It is represented by the symbol \(e\), where \(0 \leq e < 1\).
A greater eccentricity value indicates a more elongated ellipse.
The eccentricity can be determined using the formula: \( e = \sqrt{1 - \frac{a^2}{b^2}} \).
Here, \(a^2 = 0.5\) and \(b^2 = 4\), making the eccentricity \(e = \sqrt{0.875}\), approximately \(0.935\).
This relatively high value confirms our ellipse is quite elongated, with a more pronounced stretch along its major axis.

Major and Minor Axes

The major and minor axes of an ellipse are fundamental to its geometry, representing the longest and shortest diameters, respectively.
The major axis passes through the foci and the vertices, being the axis along which the ellipse is longest.
For our given ellipse equation, \( \frac{x^2}{0.5} + \frac{y^2}{4} = 1 \), the major axis is vertical because the larger denominator is beneath \(y^2\).
The length of the major axis is given by \(2b\).
Since \(b = 2\), the major axis length is \(4\).
The minor axis, being perpendicular to the major axis, is horizontal in this case.
Its length is represented by \(2a\).
Here, \(a = \frac{\sqrt{2}}{2}\), giving a minor axis length of \(\sqrt{2}\).
This information allows us to fully comprehend the size and orientation of the ellipse.

Graphing Ellipses

Graphing an ellipse requires plotting it based on its center, vertices, foci, and axes lengths.
Our ellipse is symmetrically centered at \((0, 0)\).
To graph it correctly:

• Begin by marking the center at the origin.
• Plot the vertices at \((0, \pm 2)\).
• Draw the minor axis endpoints, at roughly \((\pm \frac{\sqrt{2}}{2}, 0)\).
• Mark the foci points at \((0, \pm 1.87)\); these indicate the concentration of the ellipse's shape.

The ellipse should extend vertically more than horizontally due to its vertical major axis.
Connecting these points will depict the elongated rounded shape characteristic of an ellipse.