Book cover for Algebra and Trigonometry

Algebra and Trigonometry

Judith A. Beecher, Judith A. Penna, Marvin L. Bittinger

ISBN #9780321693983

4th Edition

5,839 Questions

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199,024 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

Parabolas are fundamental curves in analytic geometry, defined as the set of points equidistant from a focus and a directrix. The standard equations of parabolas, whether they open vertically or horizontally, allow us to immediately identify key features like the vertex, focus, and directrix. Mastering techniques such as completing the square and applying the distance formula is essential not only for drawing accurate graphs but also for solving applied problems in optics and engineering.

Learning Objectives

1

Understand the geometric definition of a parabola and identify its key components: focus, directrix, vertex, and axis of symmetry.

2

Derive and interpret the standard equations of parabolas (both vertical and horizontal) with the vertex at the origin and after translations.

3

Use the distance formula and the concept of completing the square to transform quadratic equations into standard form.

4

Graph parabolas given in standard or general form by identifying the vertex, focus, and directrix.

5

Apply the properties of parabolas to solve real-world problems in technology, optics, and engineering.

Key Concepts

CONCEPT

DEFINITION

Parabola

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

Focus

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

Directrix

A fixed line not passing through the parabola such that the distance from any point on the parabola to the directrix equals its distance to the focus.

Vertex

The point on the parabola that is equidistant from the focus and the directrix, and where the parabola changes direction.

Axis of Symmetry

The line that passes through the vertex and focus, dividing the parabola into two mirror-image halves; it is perpendicular to the directrix.

Standard Equation of a Parabola

An algebraic expression representing the parabola’s shape. For vertical parabolas: y^2 = 4px (or x^2 = 4py for horizontal ones), where p is the distance from the vertex to the focus.

Example Problems

Example 1

Match the equation with one of the graphs $(a)-(f),$ which follow. A.(Graph cannot copy) B.(Graph cannot copy) C.(Graph cannot copy) D.(Graph cannot copy) E.(Graph cannot copy) F.(Graph cannot copy) $$x^{2}=8 y$$

Example 2

Match the equation with one of the graphs $(a)-(f),$ which follow. A.(Graph cannot copy) B.(Graph cannot copy) C.(Graph cannot copy) D.(Graph cannot copy) E.(Graph cannot copy) F.(Graph cannot copy) $$y^{2}=-10 x$$

Example 3

Match the equation with one of the graphs $(a)-(f),$ which follow. A.(Graph cannot copy) B.(Graph cannot copy) C.(Graph cannot copy) D.(Graph cannot copy) E.(Graph cannot copy) F.(Graph cannot copy) $$(y-2)^{2}=-3(x+4)$$

Example 4

Match the equation with one of the graphs $(a)-(f),$ which follow. A.(Graph cannot copy) B.(Graph cannot copy) C.(Graph cannot copy) D.(Graph cannot copy) E.(Graph cannot copy) F.(Graph cannot copy) $$(x+1)^{2}=5(y-2)$$

Example 5

Match the equation with one of the graphs $(a)-(f),$ which follow. A.(Graph cannot copy) B.(Graph cannot copy) C.(Graph cannot copy) D.(Graph cannot copy) E.(Graph cannot copy) F.(Graph cannot copy) $$13 x^{2}-8 y-9=0$$
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Step-by-Step Explanations

QUESTION

How do you derive the standard equation of a parabola with vertex at the origin and focus at (0, p)?

STEP-BY-STEP ANSWER:

Step 1: Start with the definition: a parabola is the set of all points (x, y) equidistant from the focus (0, p) and the directrix y = -p.
Step 2: Write the distance from a point (x, y) to the focus using the distance formula: √(x^2 + (y - p)^2).
Step 3: Write the distance from (x, y) to the directrix, which is the vertical distance: |y + p|.
Step 4: Set the two distances equal: √(x^2 + (y - p)^2) = |y + p|.
Step 5: Square both sides to eliminate the square root: x^2 + (y - p)^2 = (y + p)^2.
Step 6: Expand both sides and simplify. The y^2 terms cancel and you are left with an equation that can be rearranged into the standard form.
Final Answer: The resulting equation takes the form x^2 = 4py if the parabola opens upward (and a corresponding form if it opens downward).

Deriving the Standard Equation from the Geometric Definition

QUESTION

How do you convert the quadratic equation x^2 + 6x + 4y + 5 = 0 into standard form and find its vertex, focus, and directrix?

STEP-BY-STEP ANSWER:

Step 1: Rearrange the equation grouping the x-terms together: x^2 + 6x = -4y - 5.
Step 2: Complete the square for x by adding (6/2)^2 = 9 to both sides: x^2 + 6x + 9 = -4y - 5 + 9.
Step 3: Write the left side as a perfect square: (x + 3)^2 = -4y + 4.
Step 4: Rearranging gives (x + 3)^2 = -4(y - 1).
Step 5: Identify the vertex as (-3, 1) and compare with the standard form (x - h)^2 = 4p(y - k) where 4p = -4.
Step 6: Solve for p: 4p = -4, so p = -1. This indicates the parabola opens downward.
Step 7: The focus is at (h, k + p) = (-3, 1 - 1) = (-3, 0) and the directrix is the line y = k - p = 1 - (-1) = 2.
Final Answer: The parabola in standard form is (x + 3)^2 = -4(y - 1) with vertex (-3, 1), focus (-3, 0), and directrix y = 2.

Completing the Square to Find the Vertex, Focus, and Directrix

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Common Mistakes

  • Confusing the direction in which the parabola opens by misinterpreting the sign of p.
  • Errors in completing the square, such as neglecting to balance the equation when adding a constant term.
  • Mixing up the roles of the x and y variables when dealing with horizontally versus vertically oriented parabolas.
  • Incorrect application of the distance formula, especially when handling absolute values.
  • Forgetting that the vertex lies exactly midway between the focus and the directrix.