Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes. \(16 x^{2}+4 y^{2}=32\)

Step-by-Step Solution

Answer

The graph is a vertical ellipse with vertices at (0, ±2√2) and foci at (0, ±√6).

Step 1: Rewrite the Equation in Standard Form

Given the equation \(16x^2 + 4y^2 = 32\). Divide the entire equation by 32 to rewrite it in standard form for an ellipse.\[\frac{x^2}{2} + \frac{y^2}{8} = 1\]

Step 2: Identify the Type of Conic

The standard form obtained is \(\frac{x^2}{2} + \frac{y^2}{8} = 1\), which is the equation of an ellipse. In the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the larger denominator under \(y^2\) indicates that the major axis is vertical.

Step 3: Determine the Vertices

The equation \(\frac{x^2}{2} + \frac{y^2}{8} = 1\) implies the semi-major axis \(b = \sqrt{8} = 2\sqrt{2}\) and semi-minor axis \(a = \sqrt{2}\). The vertices along the y-axis are \((0, \pm 2\sqrt{2})\).

Step 4: Find the Foci

The foci of the ellipse are located at \((0, \pm c)\), where \(c^2 = b^2 - a^2\). Calculate \(c\) by finding \(c = \sqrt{b^2 - a^2} = \sqrt{8 - 2} = \sqrt{6}\). So, the foci are \((0, \pm \sqrt{6})\).

Step 5: Identify Any Asymptotes (For Hyperbolas)

Since this is an ellipse, it does not have any asymptotes. Asymptotes are characteristic of hyperbolas. Thus, for ellipses, we only focus on vertices and foci.

Step 6: Sketch the Graph

Draw an ellipse centered at the origin (0,0) on a coordinate plane. Mark the vertices \((0, 2\sqrt{2}), (0, -2\sqrt{2})\) and the foci \((0, \sqrt{6}), (0, -\sqrt{6})\) along the y-axis. The ellipse is vertical due to the larger denominator under \(y^2\).

Key Concepts

Vertices of Ellipse

The vertices of an ellipse are significant points that help determine its shape and size. For any ellipse, these are the points that lie the furthest away on the major axis.
To find the vertices, we need to know which axis is major. In our example, given the equation \( \frac{x^2}{2} + \frac{y^2}{8} = 1 \), the major axis is vertical because the denominator under \(y^2\) is larger. This tells us the major axis runs along the y-axis.

• Calculate semi-major axis: \( b = \sqrt{8} = 2\sqrt{2} \)
• Calculate semi-minor axis: \( a = \sqrt{2} \)
• Find the vertices: Since the major axis is vertical, the vertices are at \((0, \pm 2\sqrt{2})\).

These vertices help us sketch the ellipse, showing its full extent vertically on the coordinate plane.

Foci of Ellipse

The foci of an ellipse are two points located inside the ellipse whose constant sum of distances to any point on the ellipse define the shape. These are essential in understanding the path and stretch of the ellipse.
For the given ellipse, we have the standard form equation \( \frac{x^2}{2} + \frac{y^2}{8} = 1 \), which already indicates a vertical orientation.

• We calculate the distance to the foci using the equation \( c^2 = b^2 - a^2 \).
• Substitute the values: \( c = \sqrt{b^2 - a^2} = \sqrt{8 - 2} = \sqrt{6} \).
• Thus, the foci are at the points \((0, \pm \sqrt{6})\) on the y-axis.

The foci are essential for graphing, as they ensure an accurate depiction of the ellipse's shape.

Standard Form of Ellipse

The standard form of an ellipse provides a concise representation of its geometric properties. This form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the lengths of the semi-minor and semi-major axes respectively.
In our problem, the ellipse equation is initially given as \( 16x^2 + 4y^2 = 32 \). This needs to be converted into the standard form to facilitate analysis and graphing.

• Divide each term by 32: \( \frac{x^2}{2} + \frac{y^2}{8} = 1 \).
• Identify \( a \) and \( b \): Here, \( a = \sqrt{2} \) and \( b = \sqrt{8} = 2\sqrt{2} \).
• This tells us the ellipse's major axis is along the y-direction, reflecting our earlier finds regarding vertices and foci.

The standard form is invaluable for recognizing and utilizing the properties of an ellipse effectively in calculations and graphing.