Problem 15 - Precalculus: Mathematical for Calculus

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: (0, ±√2); Foci: (0, ±√1.5); Eccentricity: √3/2; Major axis: 2√2, Minor axis: 2√0.5.

Step 1: Write 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 get rid of the fraction: \(2x^2 + \frac{1}{2}y^2 = 1\). Divide every term by 1 to create an ellipse equation in standard form: \(\frac{x^2}{0.5} + \frac{y^2}{2} = 1\).

Step 2: Identify Center, Vertices, and Foci

The standard form of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(b^2 > a^2\) because \(\frac{1}{2} > \frac{1}{0.5}\). Thus, \(a^2 = 0.5\) and \(b^2 = 2\), so \(a = \sqrt{0.5}\) and \(b = \sqrt{2}\). This means the ellipse's major axis is vertical. The vertices are at \((0, \pm b) = (0, \pm \sqrt{2})\). Calculate the foci using \(c^2 = b^2 - a^2\), so \(c = \sqrt{2 - 0.5} = \sqrt{1.5}\). The foci are \((0, \pm \sqrt{1.5})\).

Step 3: Determine Eccentricity

Eccentricity \(e\) is calculated using \(e = \frac{c}{b}\). Here, \(c = \sqrt{1.5}\) and \(b = \sqrt{2}\). Therefore, \(e = \frac{\sqrt{1.5}}{\sqrt{2}}\). Simplify this to get \(e = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).

Step 4: Calculate Lengths of Axes

The length of the major axis is \(2b = 2\sqrt{2}\), while the length of the minor axis is \(2a = 2\sqrt{0.5}\).

Step 5: Sketch the Graph

The ellipse is centered at the origin, with a vertical major axis of \(2\sqrt{2}\) and a horizontal minor axis of \(2\sqrt{0.5}\). The vertices are \((0, \pm \sqrt{2})\) and foci at \((0, \pm \sqrt{1.5})\). Based on these dimensions, sketch an elongated ellipse vertically centered at (0,0).

Key Concepts

Vertices of Ellipse

When you think of the vertices of an ellipse, imagine the most stretched parts of this smooth, oval shape. They mark the endpoints of the longest distance across the ellipse, known as the major axis. In a vertically oriented ellipse—like the one from the problem—the vertices are along the y-axis.

For our example, the equation has been simplified to \(\frac{x^2}{0.5} + \frac{y^2}{2} = 1\). Here, \(b^2\) is greater than \(a^2\), indicating the major axis is vertical, with \(a^2 = 0.5\) and \(b^2 = 2\). Consequently, the vertices are located at \((0, \pm \sqrt{2})\). These points create the farthest reach of the ellipse along the vertical direction.

To measure these vertices:

• Find \(b\) by taking the square root of \(b^2\). Here, \(\sqrt{2}\).
• List the vertices: \((0, \sqrt{2})\) and \((0, -\sqrt{2})\).

Understanding the vertices helps in visualizing the ellipse's size and orientation.

Foci of Ellipse

The foci of an ellipse are crucial points that define its shape. They lie along the major axis, inside the ellipse, and significantly impact its curvature. Foci are the points you would "tie" to create an ellipse by swinging a pencil around them.

For the ellipse \(\frac{x^2}{0.5} + \frac{y^2}{2} = 1\), the foci are determined using the relationship \(c^2 = b^2 - a^2\). Calculate \(c\) as follows:

• \(c^2 = 2 - 0.5 = 1.5\)
• \(c = \sqrt{1.5}\)

The foci are then at the points \((0, \pm \sqrt{1.5})\), positioned along the y-axis. Placing them correctly reveals the direction and degree of the ellipse's elongation, drawing its distinctive oval shape closer to the foci as compared to a circle.

Eccentricity of Ellipse

Eccentricity, denoted as \(e\), quantifies the 'ovalness' or how stretched out an ellipse is. It ranges between 0 (a perfect circle) and nearly 1 (a highly elongated shape). The eccentricity offers a deep dive into the ellipse’s angular properties.

In our ellipse, calculate \(e\) using \(c\) and \(b\):

• \(e = \frac{c}{b}\)
• Where \(c = \sqrt{1.5}\) and \(b = \sqrt{2}\)
• Thus, \(e = \frac{\sqrt{1.5}}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\)

With an eccentricity of \(\frac{\sqrt{3}}{2}\), our ellipse shows a distinctive lengthwise stretch compared to height. The value of \(e\) provides insight into the geometry of the ellipse, indicating how close it is to being a perfect circle.

Axes of Ellipse

Axes of an ellipse define its primary spans—the major and minor axes—and are central to understanding its shape. The major axis is the longest diameter, while the minor axis is the shortest. Their orientation indicates the general direction of the ellipse's stretch.

For our ellipse \(\frac{x^2}{0.5} + \frac{y^2}{2} = 1\):

• The major axis is vertical because \(b^2 > a^2\).
• Length of Major Axis: \(2b = 2\sqrt{2}\)
• Length of Minor Axis: \(2a = 2\sqrt{0.5}\)

The measurements tell us the full diameter across both axes. In this setup, you'll sketch an elongated ellipse with a towering vertical shape that spans \(2\sqrt{2}\) units high and \(2\sqrt{0.5}\) units wide. Understanding these axes helps effectively map and draw an ellipse, describing its fullest reach.