Problem 45
Question
a. What kind of conic section is the curve \(x y+2 x-y=0 ?\) b. Solve the equation \(x y+2 x-y=0\) for \(y\) and sketch the curve as the graph of a rational function of \(x\) . c. Find equations for the lines parallel to the line \(y=-2 x\) that are normal to the curve. Add the lines to your sketch.
Step-by-Step Solution
Verified Answer
a) Hyperbola. b) \( y = \frac{-2x}{x-1} \); sketch as rational function. c) Find lines normal to the curve parallel to \( y = -2x \) and add to sketch.
1Step 1: Recognize the Type of Conic Section
The given equation is \( x y + 2x - y = 0 \). This is a general form of a conic section, but it needs simplification. The equation represents a hyperbola because the terms in \( x \) and \( y \) are not squared and are multiplied together.
2Step 2: Solve the Equation for y
Rewrite the given equation: \[ x y + 2x - y = 0 \]Isolate \( y \): \[ y(x - 1) = -2x \] Divide both sides by \( x - 1 \): \[ y = \frac{-2x}{x - 1} \]This is the equation of \( y \) as a function of \( x \).
3Step 3: Sketch the Rational Function
The equation \( y = \frac{-2x}{x - 1} \) is a rational function. To sketch it, identify vertical and horizontal asymptotes.- Vertical Asymptote: Set denominator \( x - 1 = 0 \), so vertical asymptote is at \( x = 1 \).- Horizontal Asymptote: For large \( x \), \( y \approx -2 \), so the horizontal asymptote is \( y = -2 \).Sketch starts from these asymptotes.
4Step 4: Find the Normal Lines
Normal lines to the curve are perpendicular to the tangent lines at the points of the curve, but the problem specifies using lines parallel to \( y = -2x \).Normal lines have slopes of the negative reciprocal of the tangent lines. The slopes of the tangent lines equal the derivative of \( y = \frac{-2x}{x - 1} \). Calculate the derivative and find points: \[ y' = \frac{2}{(x - 1)^2} \]Since we need lines parallel to \( y = -2x \), we set this slope \( -2 = \frac{2}{(x - 1)^2} \), solved for points on the curve.
5Step 5: Draw and Add Lines to Sketch
Using the solutions from Step 4, add the lines parallel to \( y = -2x \) that intersect the curve.Sketch these lines on the same graph as the curve, considering the asymptotes found in Step 3. If solutions exist, hence there will be intersections of these lines on the sketch.
Key Concepts
HyperbolaRational FunctionsAsymptotesNormal Lines
Hyperbola
A hyperbola is a type of conic section formed when a plane intersects both nappes of a double cone. Unlike ellipses and circles, hyperbolas consist of two separate, symmetrical curves known as branches. In the context we have here, the equation \( xy + 2x - y = 0 \) represents a hyperbola because it includes linear terms in both \( x \) and \( y \) multiplied together, without involving square powers.
- Hyperbolas are defined by their asymptotes which are straight lines that the hyperbola approaches but never actually meets.
- Understanding the nature of the hyperbola helps you predict its shape and behavior across different values of \( x \) and \( y \).
Rational Functions
Rational functions are those of the form \( f(x) = \frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials. In the equation \( y = \frac{-2x}{x - 1} \), \( p(x) = -2x \) and \( q(x) = x-1 \). This is a simple rational function because both \( p(x) \) and \( q(x) \) are linear polynomials.
- Such functions are not defined at points where the denominator equals zero, leading to vertical asymptotes.
- For rational functions, horizontal or oblique asymptotes can also be determined based on the degrees of the numerator and denominator.
Asymptotes
Asymptotes are critical lines that a curve approaches as it heads towards infinity. They are fundamental when analyzing rational functions and conics such as hyperbolas.
- Vertical Asymptotes occur where the function's denominator is zero, leading to an undefined point—in our case, \( x = 1 \).
- Horizontal Asymptotes show the behavior of a function as \( x \) approaches large positive or negative values. Here, as \( x \) increases, \( y \) tends towards \( -2 \).
Normal Lines
Normal lines to a curve are perpendicular to the tangent lines at a given point on the curve. To find these for a rational function like \( y = \frac{-2x}{x - 1} \), one would first calculate the derivative, which gives the slope of the tangent.
- Once the derivative is determined (\( y' = \frac{2}{(x - 1)^2} \)), calculate the negative reciprocal for the slope of the normal line.
- In this exercise, lines parallel to \( y = -2x \) are considered, which implies setting the slope of the normal line equal to -2.
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