Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the hyperbolas \(x y=1\) and \(x^{2}-y^{2}=1\) intersect at right angles.

Step-by-Step Solution

Answer

The hyperbolas intersect at the points: \((\pm \sqrt{\frac{1 + \sqrt{5}}{2}}, \pm \sqrt{\frac{2}{1 + \sqrt{5}}})\), and their tangents show perpendicular slopes.

Step 1: Find the Points of Intersection

To find the points where the hyperbolas intersect, solve the system of equations given by the hyperbolas. The equations are: \[ xy = 1 \]\[ x^2 - y^2 = 1 \]From the first equation, express \( y \) in terms of \( x \):\[ y = \frac{1}{x} \]Substitute this expression into the second equation:\[ x^2 - \left(\frac{1}{x}\right)^2 = 1 \]Simplify and solve for \( x \):\[ x^2 - \frac{1}{x^2} = 1 \]Multiply through by \( x^2 \) to eliminate the fraction:\[ x^4 - 1 = x^2 \]Rearrange to form a quadratic in \( x^2 \):\[ x^4 - x^2 - 1 = 0 \]Let \( z = x^2 \), thus:\[ z^2 - z - 1 = 0 \]Solve this quadratic equation using the quadratic formula:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = -1, c = -1 \):\[ z = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]So, \( x^2 = \frac{1 + \sqrt{5}}{2} \) or \( x^2 = \frac{1 - \sqrt{5}}{2} \). The second solution is not valid as it would yield a negative squared term.

Step 2: Solve for Corresponding Values of y

Once \( x^2 = \frac{1 + \sqrt{5}}{2} \), find \( x \) by taking the square root. There are two solutions for \( x \):\[ x = \pm \sqrt{\frac{1 + \sqrt{5}}{2}} \]Since \( y = \frac{1}{x} \), the corresponding values of \( y \) are:\[ y = \pm \sqrt{\frac{2}{1 + \sqrt{5}}} \]So the intersection points are:\( \left( \sqrt{\frac{1 + \sqrt{5}}{2}}, \sqrt{\frac{2}{1 + \sqrt{5}}} \right) \), \( \left( -\sqrt{\frac{1 + \sqrt{5}}{2}}, -\sqrt{\frac{2}{1 + \sqrt{5}}} \right) \).

Step 3: Determine Slopes at the Intersection Points

Find the slopes of the tangent lines to each curve at the points of intersection. For the hyperbola \( xy = 1 \), differentiate implicitly:\[ x\frac{dy}{dx} + y = 0 \] Thus, the slope \( \frac{dy}{dx} \) is:\[ \frac{dy}{dx} = -\frac{y}{x} \]For \( x^2 - y^2 = 1 \), differentiate:\[ 2x - 2y\frac{dy}{dx} = 0 \] so:\[ \frac{dy}{dx} = \frac{x}{y} \]Therefore, at any intersection point, the slopes are \( -\frac{y}{x} \) and \( \frac{x}{y} \).

Step 4: Check if Tangents are Perpendicular

To verify if the curves intersect at right angles, check if the product of their slopes at the intersection points equals \(-1\). Multiply the slopes calculated in the previous step:\[ \left( -\frac{y}{x} \right) \times \left( \frac{x}{y} \right) = -1 \]This shows that the tangents are perpendicular, confirming the hyperbolas intersect at right angles.

Key Concepts

Implicit Differentiation

Implicit differentiation is a method used in calculus when dealing with equations that define a relationship between variables without solving for one variable in terms of the other. When we have an equation like \( xy = 1 \), neither \( x \) nor \( y \) is explicitly isolated. To find the slope of the tangent line to this curve, we treat \( y \) as a function of \( x \), denoted \( y(x) \), and differentiate both sides of the equation with respect to \( x \).During implicit differentiation, remember to use the chain rule. For the equation \( xy = 1 \), differentiate as follows: apply the product rule to \( xy \), getting \( x(\frac{dy}{dx}) + y = 0 \). Solving for \( \frac{dy}{dx} \), we find the slope of the tangent line is \(-\frac{y}{x}\), which is an essential step to determine the angle of intersection between hyperbolas.

System of Equations

A system of equations involves solving two or more equations simultaneously. In our case, the hyperbolas are represented by the system:

• \( xy = 1 \)
• \( x^2 - y^2 = 1 \)

To find the intersection points of these hyperbolas, we need to solve this system. We start by expressing one variable in terms of the other from one equation, like \( y = \frac{1}{x} \) from the first equation, and substitute into the second equation.This substitution transforms the system into a single equation in one variable, which can then be solved by applying algebraic manipulations and techniques such as the quadratic formula, which we'll discuss further in another section. This method allows us to determine the points where the hyperbolas intersect.

Right Angle

When two curves intersect at a right angle, the tangent lines at the point of intersection are perpendicular. The definition of perpendicular lines in terms of slopes is that their product is \(-1\). This concept is crucial to demonstrate the right-angle intersection of the hyperbolas given by \( xy = 1 \) and \( x^2 - y^2 = 1 \).After finding the slopes of the tangent lines from implicit differentiation, \(-\frac{y}{x}\) and \(\frac{x}{y}\), check for perpendicularity. Calculate the product:

• \[-\frac{y}{x} \times \frac{x}{y} = -1\]

This confirms the tangents indeed meet at a right angle, proving our geometric assertion about the intersection of these hyperbolas.

Quadratic Formula

Quadratic equations are often encountered while solving a system of equations. In our specific exercise, the substitution led to a quadratic equation \( z^2 - z - 1 = 0 \) where \( z = x^2 \).The quadratic formula provides the solutions for the equation \( az^2 + bz + c = 0 \) as:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation, \( a = 1, b = -1, c = -1 \). Substituting these values gives us \[ z = \frac{1 \pm \sqrt{5}}{2} \].This formula is powerful as it provides solutions even when short of simple factorization, making it indispensable for finding intersection points and verifying the right-angle intersection in mathematical problems involving quadratic equations.