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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.7.52
Chapter 3, Problem 3.7.52

Normal lines to a parabola Show that if it is possible to draw three normal lines from the point (a, 0) to the parabola x = y² shown in the accompanying diagram, then a must be greater than 1/2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular?


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To find the normal lines to the parabola x = y² from the point (a, 0), we first need to determine the slope of the tangent line to the parabola at any point (x, y). For the parabola x = y², differentiate implicitly with respect to y to find the slope of the tangent line: dx/dy = 2y.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/(2y).
The equation of the normal line passing through the point (x₀, y₀) on the parabola and the point (a, 0) is given by: y - y₀ = -1/(2y₀) * (x - x₀).
Substitute x₀ = y₀² into the normal line equation and set y = 0 to find the x-intercept, which is the point (a, 0). This gives: 0 - y₀ = -1/(2y₀) * (a - y₀²).
Solve the resulting equation for a in terms of y₀. Analyze the conditions under which this equation has three solutions for y₀, which correspond to three normal lines. Show that a must be greater than 1/2 for three solutions to exist, and determine the value of a for which the other two normal lines are perpendicular by setting their slopes to be negative reciprocals of each other.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Line

A normal line to a curve at a given point is a line that is perpendicular to the tangent line at that point. For a parabola, the slope of the normal line can be determined using the derivative of the function. Understanding how to find the slope of the tangent line is crucial, as the slope of the normal line is the negative reciprocal of this slope.
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Slopes of Tangent Lines

Parabola Properties

A parabola is a symmetric curve defined by a quadratic equation, typically in the form y = ax² + bx + c. The specific parabola in the question is given by x = y², which opens to the right. Recognizing the properties of parabolas, such as their vertex and axis of symmetry, is essential for analyzing normal lines and their intersections with points outside the curve.
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Properties of Parabolas

Perpendicular Lines

Two lines are perpendicular if the product of their slopes is -1. In the context of the problem, finding the value of 'a' such that two normal lines are perpendicular involves setting up an equation based on their slopes. This concept is fundamental in determining the conditions under which multiple normal lines can exist from a single point to a curve.
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Slopes of Tangent Lines
Related Practice
Textbook Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


xy + 2x - 5y = 2, (3, 2)

317
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Textbook Question

Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (−π/2, π/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.

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Textbook Question

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x²(x – y)² = x² – y²

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Textbook Question

In Exercises 29 and 30, find the slope of the curve at the given points.


y² + x² = y⁴ – 2x at (–2,1) and (–2,–1)

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Textbook Question

For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.


lim (x → −1) (x²/⁹ − 1) / (x + 1)

202
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Textbook Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


x³/² + 2y³/² = 17, (1, 4)

217
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