Question

Suppose two hyperbolas with eccentricities \(e\) and \(E\) have perpendicular major axes and share a set of asymptotes. Show that \(e^{-2}+E^{-2}=1\).

Short answer

Question: Given two hyperbolas with perpendicular major axes and sharing a set of asymptotes, prove that \(e^{-2}+E^{-2}=1\), where \(e\) and \(E\) are their eccentricities. Solution: Following the steps outlined in the provided solution, we derived the expressions for \(e^{-2}\) and \(E^{-2}\) based on the definition of eccentricity for hyperbolas, and then used the property of common asymptotes to simplify and combine the expressions. The result is \(e^{-2} + E^{-2} = 1\).

Step-by-step solution

Define the hyperbolas

Let the equations of the two hyperbolas be in Cartesian coordinates: Hyperbola 1: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) Hyperbola 2: \(\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1\) Where \(a\), \(b\), \(A\), and \(B\) are the semi-major and semi-minor axes of the respective hyperbolas.

Determine the asymptotes for each hyperbola

For hyperbola 1, the asymptotes are given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0\) or \(\frac{x^2}{a^2} =\frac{y^2}{b^2}\). Similarly, for hyperbola 2, the asymptotes are given by the equation \(\frac{y^2}{A^2} - \frac{x^2}{B^2} = 0\) or \(\frac{y^2}{A^2} =\frac{x^2}{B^2}\).

Use the common asymptotes

Since both hyperbolas share a set of common asymptotes, we can set the asymptote equations equal to each other: \(\frac{x^2}{a^2} = \frac{y^2}{b^2} = \frac{y^2}{A^2} = \frac{x^2}{B^2}\).

Utilize the definition of eccentricity for hyperbolas

For a hyperbola, eccentricity is given by the equation \(e =\sqrt{1 + \frac{b^2}{a^2}}\). For Hyperbola 1, we have \(e^2 = 1 + \frac{b^2}{a^2}\) and for Hyperbola 2, \(E^2 = 1 + \frac{B^2}{A^2}\).

Derive expressions for \(e^{-2}\) and \(E^{-2}\)

We can rearrange the eccentricity equations from Step 4 and find the expressions for \(e^{-2}\) and \(E^{-2}\). For hyperbola 1, we have \(e^{-2} = 1 - \frac{b^2}{a^2}\). For hyperbola 2, we have \(E^{-2} = 1 - \frac{B^2}{A^2}\).

Add \(e^{-2}\) and \(E^{-2}\) together

Now we simply add the expressions \(e^{-2}\) and \(E^{-2}\): \(e^{-2} + E^{-2} = \left(1 - \frac{b^2}{a^2}\right) + \left(1 - \frac{B^2}{A^2}\right)\).

Use the common asymptotes to simplify

From Step 3, we have \(\frac{b^2}{a^2} = \frac{x^2}{y^2}\) and \(\frac{B^2}{A^2} = \frac{y^2}{x^2}\). Using these expressions, we can simplify our equation from Step 6: \(e^{-2} + E^{-2} = \left(1 - \frac{x^2}{y^2}\right) + \left(1 - \frac{y^2}{x^2}\right)\). By adding the two expressions together, we get: \(e^{-2} + E^{-2} = 1\). Hence, the equation \(e^{-2} + E^{-2} = 1\) holds true for the given condition of two hyperbolas with perpendicular major axes and sharing a set of asymptotes.

Key concepts

Eccentricity of Hyperbolas

The eccentricity of a hyperbola is a key concept in understanding its shape and properties. Mathematically, the eccentricity (\( e \) for hyperbola 1 and \( E \) for hyperbola 2 in our exercise) is defined as the ratio of the distance from any point on the hyperbola to one of its foci, and the distance from that point to a corresponding directrix.

For hyperbolas, the eccentricity is always greater than 1, which distinguishes them from ellipses (where the eccentricity is less than 1). It can be calculated using the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \) for each hyperbola, where \( a \) is the length of the semi-major axis and \( b \) is the length of the semi-minor axis.

The higher the eccentricity, the more 'stretched' the hyperbola looks. In the given problem, we used the concept of eccentricity inverses, \( e^{-2} \) and \( E^{-2} \) to show a unique relationship when hyperbolas share asymptotes and have perpendicular major axes, leading to the elegant identity \( e^{-2} + E^{-2} = 1 \) that encompasses the inherent symmetry of their asymptotic behavior.

Major Axes of Hyperbolas

Every hyperbola features two major axes which are the primary reference lines from which its shape and equation are defined. The major axis of a hyperbola is the segment that runs through the foci, with its midpoint at the center of the hyperbola. The major axis is essentially the longest straight line that can be drawn through the center of the hyperbola, and it is always associated with the larger denominator in the hyperbola’s standard equation.

In the Cartesian coordinate system, the axes of the hyperbola can be either horizontal or vertical. For a horizontal major axis, the standard form of the equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) while for a vertical major axis, it flips to \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) where \( a \) and \( b \) are as previously mentioned, the lengths of the semi-major and semi-minor axes, respectively.

In the exercise, the perpendicular orientation of major axes signifies a fundamental geometric relationship that contributes to the derivation of the formula linking the eccentricities.

Cartesian Coordinates System

The Cartesian coordinates system is a cornerstone of modern geometry and a crucial framework for understanding hyperbolas and their properties. It is composed of two perpendicular lines, known as axes, typically labeled as X (horizontal) and Y (vertical), that intersect at a point called the origin. The position of any point in this plane is determined by two numbers, its X-coordinate and Y-coordinate.

Hyperbolas, like our exercise’s examples, are often represented in this system using their standard equation forms, which make it relatively straightforward to graph them as well as to identify key features such as vertices, foci, and asymptotes.

The exercise required us to work in the Cartesian coordinates system to prove a relationship about hyperbolas sharing a set of asymptotes and having perpendicular major axes. Understanding this system and how it relates to geometric figures such as hyperbolas is essential for comprehending their characteristics and for solving complex algebraic problems.