Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes. \(\frac{x^{2}}{7}+\frac{y^{2}}{4}=1\)

Step-by-Step Solution

Answer

The graph is a horizontal ellipse with vertices at \((\pm \sqrt{7}, 0)\) and foci at \((\pm \sqrt{3}, 0)\); no asymptotes.

Step 1: Identify the Conic Section

The given equation is \( \frac{x^2}{7} + \frac{y^2}{4} = 1 \). This is the standard form of the equation of an ellipse centered at the origin (0, 0), where the terms \( x^2 \) and \( y^2 \) are added.

Step 2: Identify the Axes Lengths

Since the equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), and the larger denominator \( 7 \) is under \( x^2 \), the ellipse is horizontal. Hence, \( a^2 = 7 \) and \( b^2 = 4 \). Calculate \( a \) and \( b \): \( a = \sqrt{7} \) (approximately 2.65) and \( b = \sqrt{4} = 2 \).

Step 3: Determine the Vertices

The vertices of the ellipse are located at the endpoints of the major axis. Since the ellipse is horizontal, the vertices are \((\pm a, 0)\). Therefore, the vertices are \( (\pm \sqrt{7}, 0) \), approximately \((\pm 2.65, 0)\).

Step 4: Calculate the Foci

The foci are found using the relation \( c^2 = a^2 - b^2 \). Compute \( c = \sqrt{7 - 4} = \sqrt{3} \), approximately 1.73. The foci are \((\pm \sqrt{3}, 0)\), approximately \((\pm 1.73, 0)\).

Step 5: Discuss Asymptotes

Since this is an ellipse, it does not have asymptotes. Asymptotes are characteristic of hyperbolas, not ellipses.

Step 6: Sketch the Graph

Draw a horizontal ellipse centered at the origin. Mark the vertices at \((\pm \sqrt{7}, 0)\) and the foci at \((\pm \sqrt{3}, 0)\). The endpoints of the minor axis are at \((0, \pm 2)\). This completes the sketch of the ellipse.

Key Concepts

Ellipses

An ellipse is a type of conic section or plane curve that outlines the locus of all points for which the sum of the distances to two focal points (foci) is constant. It's an elongated circle stretched along one axis, forming an oval shape. Its symmetry and smooth curve make ellipses a familiar shape in mathematics. Ellipses occur naturally in various phenomena including planetary orbits in astronomy. They are characterized by two main axes: the major axis, which is the longest diameter of the ellipse, and the minor axis, which is the shortest.

• The major axis runs through the center and both foci, determining the ellipse's longest extent.
• The minor axis is perpendicular to the major axis and runs through the center.
• Ellipses have a property where the sum of the distances from any point on the ellipse to the two foci is constant.

In the context of mathematical equations and graphs, ellipses are identified by their standard equation, which allows us to define precise properties like the vertices and foci.

Vertices of an Ellipse

In an ellipse, the vertices are key points located at the ends of the major axis. They represent the farthest points from the center, stretching along the direction of the major axis. Vertices are crucial as they help visualize and define the shape and size of the ellipse. For an ellipse centered at the origin with the standard equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a > b\):

• The vertices are located at \((\pm a, 0)\) if the major axis is horizontal.
• If the major axis is vertical, the vertices will be at \((0, \pm a)\).
• The distance between the vertices is equal to the length of the major axis, \(2a\).

In the given equation \(\frac{x^2}{7} + \frac{y^2}{4} = 1\), the major axis is horizontal with vertices at \( (\pm \sqrt{7}, 0) \), or approximately \((\pm 2.65, 0)\), showing how wide the ellipse extends along the x-axis.

Foci of an Ellipse

The foci of an ellipse are two special points located along the major axis around which the ellipse is shaped. Unlike circles that have a single center, ellipses have two foci, contributing to their unique form. The position of the foci determines the shape of the ellipse: the closer the foci are to each other, the more the ellipse resembles a circle. For the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the foci can be found using:

• Calculate \(c\) using the formula \(c^2 = a^2 - b^2\), where \(c\) is the focal distance from the center.
• The foci are located at \((\pm c, 0)\) if the major axis is horizontal, or \((0, \pm c)\) if vertical.

For the equation \(\frac{x^2}{7} + \frac{y^2}{4} = 1\), the foci are at \((\pm \sqrt{3}, 0)\), approximately \((\pm 1.73, 0)\), highlighting their role in defining the precise shape of the ellipse.

Equation of an Ellipse

The standard equation of an ellipse is crucial for defining its geometric properties and graphing its shape. The general form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is used for ellipses centered at the origin, where \(a\) and \(b\) are the semi-major and semi-minor axis lengths, respectively:

• If \(a > b\), the ellipse is oriented horizontally.
• If \(b > a\), the ellipse is oriented vertically.

The lengths \(a\) and \(b\) are the square roots of the denominators under the \(x^2\) and \(y^2\) terms:

• For \(x^2\) divided by a larger number, the ellipse stretches along the x-axis.
• For \(y^2\) divided by a larger number, it stretches along the y-axis.

From the equation \(\frac{x^2}{7} + \frac{y^2}{4} = 1\), we identify a horizontal ellipse with semi-major axis \( a = \sqrt{7} \) and semi-minor axis \( b = \sqrt{4} = 2 \), providing a blueprint for sketching its graph.