Problem 75 - Calculus Early Transcendentals: Pearson New International Edition

Question

If a horizontal hyperbola and a vertical hyperbola have the same asymptotes, show that their eccentricities \(e\) and \(E\) satisfy \(e^{-2}+E^{-2}=1\).

Step-by-Step Solution

Answer

For horizontal and vertical hyperbolas with the same asymptotes, their eccentricities satisfy \( e^{-2} + E^{-2} = 1 \).

Step 1: Understand Hyperbola Equations

A horizontal hyperbola with center at origin has the equation \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \], whereas a vertical hyperbola with center at origin has \[ \frac{y^2}{c^2} - \frac{x^2}{d^2} = 1 \]. The asymptotes for the horizontal hyperbola are given by \( y = \pm \frac{b}{a}x \). The vertical hyperbola has asymptotes \( y = \pm \frac{c}{d}x \). Since both hyperbolas have the same asymptotes, we have \( \frac{b}{a} = \frac{c}{d} \).

Step 2: Eccentricity Definition

The eccentricity of a hyperbola is defined as \( e = \sqrt{1 + \frac{b^2}{a^2}} \) for a horizontal hyperbola and \( E = \sqrt{1 + \frac{d^2}{c^2}} \) for a vertical hyperbola. We should relate these eccentricities to the common asymptotes condition.

Step 3: Derive Relationship from Asymptotes

Using \( \frac{b}{a} = \frac{c}{d} \), we can write \( b^2d^2 = a^2c^2 \). By substituting in the expressions for \( e^2 \) and \( E^2 \), we have \( e^2 = 1 + \frac{b^2}{a^2} \) and \( E^2 = 1 + \frac{d^2}{c^2} \).

Step 4: Simplify the Expressions

Now substitute \( b^2 = a^2 \frac{b^2}{a^2} \) and \( d^2 = c^2 \frac{d^2}{c^2} \) into the respective eccentricity equations: \( e^2 - 1 = \frac{b^2}{a^2} \) and \( E^2 - 1 = \frac{d^2}{c^2} \). Correspondingly, \( \frac{b^2}{a^2} = e^2 - 1 \) and \( \frac{d^2}{c^2} = E^2 - 1 \).

Step 5: Combine and Simplify to Find the Equation

Substitute back into \( b^2d^2 = a^2c^2 \), write \( a^2(e^2 - 1)d^2 = c^2(E^2 - 1)b^2 \). Simplifying gives \( (e^2 - 1)(E^2 - 1) = 1 - e^2 - E^2 + 1 \), or \( e^{-2} + E^{-2} = 1 \).

Step 6: Conclusion: Verify Relation

After simplification, you confirm that the relationship \( e^{-2} + E^{-2} = 1 \) holds true for eccentricities of horizontal and vertical hyperbolas with the same asymptotes.

Key Concepts

Eccentricity

Eccentricity is a measure of how "stretched out" a hyperbola is, compared to a perfect circle. In mathematics, it is denoted as \( e \) for any conic section. For hyperbolas:

• A horizontal hyperbola's eccentricity is given by the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \).
• A vertical hyperbola uses \( E = \sqrt{1 + \frac{d^2}{c^2}} \).

These equations tell us how the shape of the hyperbola changes as the values of \( a, b, c, \) and \( d \) change. The eccentricity is always greater than 1 for hyperbolas, which makes them distinct from ellipses. Understanding this measure is crucial in defining the separation between the branches of a hyperbola, which is why it's so important in analyzing these curves.

Hyperbola Equations

Hyperbolas are characterized by their unique equations, which differ based on their orientation:

• Horizontal hyperbolas follow the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
• Vertical hyperbolas use \( \frac{y^2}{c^2} - \frac{x^2}{d^2} = 1 \).

In these equations, \( a, b, c, \) and \( d \) are constants that determine the shape and direction of the hyperbola.
The value of \( a \) refers to the distance from the center to the vertices along the transverse axis for horizontal hyperbolas, while \( b \) relates to the conjugate axis. For vertical hyperbolas, \( c \) and \( d \) play these roles respectively.
These equations are fundamental in understanding hyperbolas' structural characteristics and deriving further properties such as asymptotes and eccentricity.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never quite reaches. They give hyperbolas their distinct open shape. For hyperbolas:

• The horizontal hyperbola's asymptotes are \( y = \pm \frac{b}{a}x \).
• The vertical hyperbola has asymptotes \( y = \pm \frac{c}{d}x \).

The peculiarity of asymptotes for a hyperbola is that they intersect at the hyperbola's center. This center becomes a common meeting point for the diametrically opposite limbs of the hyperbola as they infinitely extend. Understanding how these asymptotes function is vital for visualizing the behavior of hyperbolas, particularly when they share asymptotes as in this problem.
Such shared conditions help us derive relationships between different components of hyperbolas, like eccentricities, as shown in the given solution.

Mathematical Proofs

Mathematical proofs provide a logical explanation of how we derive certain results, such as relationships within geometry.
In this problem, proving that \( e^{-2} + E^{-2} = 1 \) involves:

• Identifying the shared conditions between two equations, i.e., the equal asymptotes.
• Utilizing known equations for eccentricities \( e \) and \( E \) to express these conditions effectively.
• Simplifying algebraic expressions step-by-step to showcase the derived relationship as accurately as possible.

Proofs not only verify mathematical laws but also offer insight into the intrinsic connections between different geometric properties. By showing how eccentricities are related, we understand more deeply how the shape and orientation of hyperbolas are intrinsically linked, reinforcing fundamental concepts in geometry.