Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the coordinates of the foci of the ellipse that is the intersection of \(z=x^{2} / 4+y^{2} / 9\) with the plane \(z=4\).

Step-by-Step Solution

Answer

The foci are at coordinates \((0, 2\sqrt{5})\) and \((0, -2\sqrt{5})\).

Step 1: Determine the Type of Conic Section

Given the equation of a surface \( z = \frac{x^2}{4} + \frac{y^2}{9} \) and the plane \( z = 4 \), substitute \( z = 4 \) into the equation to find the equation of the ellipse or hyperbola. This gives us the equation \( 4 = \frac{x^2}{4} + \frac{y^2}{9} \). Rearranging, we get \( \frac{x^2}{4} + \frac{y^2}{9} = 4 \). Divide through by 4 to standardize: \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \). This equation represents an ellipse.

Step 2: Identify the Parameters of the Ellipse

For an ellipse in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), determine the values of \( a^2 \) and \( b^2 \). Here, \( a^2 = 16 \) and \( b^2 = 36 \), where \( a = \sqrt{16} = 4 \) and \( b = \sqrt{36} = 6 \). Since \( b > a \), this is a vertical ellipse.

Step 3: Calculate the Foci

For an ellipse, the relationship between the semi-major axis \( b \), semi-minor axis \( a \), and the distance \( c \) from the center to each focus is given by \( c = \sqrt{b^2 - a^2} \). Compute \( c \) for the vertical ellipse: \( c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \).

Step 4: Determine the Coordinates of the Foci

Since the ellipse is vertical (major axis along the y-axis), and the center is at the origin, the coordinates of the foci are \((0, +c)\) and \((0, -c)\). Substitute \( c = 2\sqrt{5} \) to get the coordinates \((0, 2\sqrt{5})\) and \((0, -2\sqrt{5})\).

Key Concepts

Conic Sections

Conic sections are the curves formed by intersecting a plane with a double-napped cone.
Each conic section represents a different type of curve, and the nature of the intersection dictates the type of conic section formed.
For example:

• If the plane cuts parallel to the base of the cone, a circle is formed.
• If the intersection angle is oblique yet does not pass through the apex, an ellipse is produced.
• A hyperbola results when the plane slices through both halves of the cone.
• A parabola is created if the plane is parallel to the side of the cone.

The problem above involves an elliptical conic section. By intersecting the given surface with a specific plane, the resulting intersection describes the equation of an ellipse, leading to investigations of its geometric properties.

Ellipse Equation

An ellipse is defined as the set of all points where the sum of the distances from two fixed points (foci) remains constant.
This problem introduces an equation for finding such an ellipse: \(\frac{x^2}{16} + \frac{y^2}{36} = 1\).
This equation is the standard form of an ellipse centered at the origin.

• The denominator of each fraction is the square of the axis length along that direction.
• If the larger denominator is under the \(y^2\) term, the ellipse is vertical. Conversely, if it's under the \(x^2\) term, the ellipse is horizontal.

Understanding this equation helps in analyzing the geometrical shape and also provides insights into its size and orientation. The values of \(a\) and \(b\) directly give the lengths of the semi-axes, vital for further calculations.

Semi-Major and Semi-Minor Axes

In the world of ellipses, the axes provide essential information about an ellipse's dimensions and orientation.
The ellipse in this context is characterized by its semi-major and semi-minor axes, which determine its shape.

• The semi-major axis is the longer of the two, measured from the center to the ellipse's edge.
• The semi-minor axis is the shorter distance from the center to the edge.

For this problem:

• The semi-major axis \(b = 6\), since the larger denominator is under \(y^2\).
• The semi-minor axis \(a = 4\), as the smaller denominator lies beneath \(x^2\).

These axes dictate the spread and position of an ellipse, guiding us to crucial points like the foci. The foci, essential for understanding the depth of an ellipse, are calculated using \(c = \sqrt{b^2 - a^2}\), which is \(c = 2\sqrt{5}\) in this exercise.