Book cover for Precalculus: A Right Triangle Approach

Precalculus: A Right Triangle Approach

Judith A. Beecher, Marvin L. Bittinger, David J. Ellenbogen, Judith A. Penna

ISBN #9780321783967

5th Edition

5,905 Questions

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25,202 Students Helped

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section delved into the geometric and algebraic properties of parabolas, emphasizing the definition based on equidistance from a focus and a directrix. Through derivations and examples, we learned how to obtain the standard form of a parabola, how to complete the square to transform quadratic equations, and how to extract key features such as the vertex, focus, and directrix. Additionally, the practical applications of parabolic shapes in technology, like satellite dishes and headlights, were explored.

Learning Objectives

1

Describe the geometric definition of a parabola, including the roles of the focus, directrix, vertex, and axis of symmetry.

2

Derive and manipulate the standard equations of parabolas with vertical and horizontal axes of symmetry.

3

Apply the method of completing the square to convert quadratic equations into standard parabola form.

4

Analyze and identify key features (vertex, focus, directrix) from a given parabola’s equation.

5

Explore real-world applications of parabolic shapes in technology and design, such as headlights and satellite dishes.

Key Concepts

CONCEPT

DEFINITION

Parabola

The set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

Focus

A fixed point located inside the parabola from which distances to any point on the parabola are equal to the distances from that point to the directrix.

Directrix

A fixed line used in the definition of a parabola such that every point on the parabola is equidistant from it and the focus.

Vertex

The point on the parabola that is exactly midway between the focus and the directrix. It is also the point where the parabola changes direction.

Axis of Symmetry

The line that passes through the focus and vertex, dividing the parabola into two mirror-image halves.

Standard Equation

An equation written in a form that clearly shows the vertex and the parameter p (distance from vertex to focus), e.g., y² = 4px for a horizontal axis or x² = 4py for a vertical axis.

Example Problems

Example 1

Match the equation with one of the graphs $(a)-(f),$ which follow. $$ x^{2}=8 y $$

Example 2

Match the equation with one of the graphs $(a)-(f),$ which follow. $$ y^{2}=-10 x $$

Example 3

Match the equation with one of the graphs $(a)-(f),$ which follow. $$ (y-2)^{2}=-3(x+4) $$

Example 4

Match the equation with one of the graphs $(a)-(f),$ which follow. $$ (x+1)^{2}=5(y-2) $$

Example 5

Match the equation with one of the graphs $(a)-(f),$ which follow. $$ 13 x^{2}-8 y-9=0 $$
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Step-by-Step Explanations

QUESTION

Given a parabola with vertex at (0,0) and focus (0,p), derive the standard form of its equation.

STEP-BY-STEP ANSWER:

Step 1: Apply the definition of a parabola: for any point P(x, y) on the parabola, the distance to the focus (0, p) equals the distance to the directrix y = -p.
Step 2: Write the distance from P(x, y) to the focus: √(x² + (y - p)²).
Step 3: Write the distance from P(x, y) to the directrix: |y + p|.
Step 4: Set the distances equal: √(x² + (y - p)²) = |y + p|.
Step 5: Square both sides to remove the square root: x² + (y - p)² = (y + p)².
Step 6: Expand both sides: x² + y² - 2py + p² = y² + 2py + p².
Step 7: Simplify by cancelling y² and p²: x² - 2py = 2py.
Step 8: Solve for y: x² = 4py.
Final Answer: The standard equation of the vertical parabola is x² = 4py.

Deriving the Standard Equation of a Vertical Parabola

QUESTION

For the parabola given by the equation x² + 6x + 4y + 5 = 0, complete the square to find the vertex, focus, and directrix.

STEP-BY-STEP ANSWER:

Step 1: Group the x-terms: x² + 6x = -4y - 5.
Step 2: Complete the square for x: Add (6/2)² = 9 to both sides to obtain x² + 6x + 9 = -4y - 5 + 9.
Step 3: Rewrite the left side as a perfect square: (x + 3)² = -4y + 4.
Step 4: Solve for y: -4y = (x + 3)² - 4, then y = -((x + 3)²)/4 + 1.
Step 5: Identify the vertex from the form y = a(x - h)² + k: Vertex is (-3, 1).
Step 6: Determine p using the coefficient relationship: In standard form, y = (1/(4p)) (x - h)² + k. Here, 1/(4p) = -1/4, so p = -1.
Step 7: Find the focus: For a vertical parabola, focus is at (h, k + p) = (-3, 1 - 1) = (-3, 0).
Step 8: Find the directrix: Line given by y = k - p = 1 - (-1) = 2.
Final Answer: Vertex: (-3, 1), Focus: (-3, 0), Directrix: y = 2.

Completing the Square to Find Parabola Features

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Common Mistakes

  • Confusing the roles of the focus and the directrix, particularly misidentifying which side of the vertex each lies on.
  • Incorrectly completing the square – forgetting to add the necessary constant to both sides of the equation.
  • Mixing up the orientation of the parabola (vertical versus horizontal), leading to errors in formulating the standard equation.
  • Overlooking sign conventions for the parameter p, especially when the parabola opens down or to the left.