Ch. 7 - Conic Sections

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

Blitzer 8th Edition

Ch. 7 - Conic Sections

Problem 3

Chapter 8, Problem 3

In Exercises 1–4, find the focus and directrix of each parabola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d). x^2 = - 4y

Verified step by step guidance

Identify the form of the given equation. The equation is

x^{2} = -4y

, which resembles the standard form of a vertical parabola:

x^{2} = 4py

Compare the given equation to the standard form to find the value of

p

. Here,

4p = -4

, so solve for

p

by dividing both sides by 4, giving

p = -1

Recall that for a parabola in the form

x^2 = 4py

, the vertex is at the origin

(0,0)

, the focus is at

(0, p)

, and the directrix is the line

y = -p

Using the value of

p = -1

, determine the focus:

(0, -1)

, and the directrix:

y = 1

Match the equation to the graph that shows a parabola opening downward (since

p

is negative), with the focus at

(0, -1)

and the directrix at

y = 1

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Key Concepts

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

Standard Form of a Parabola

A parabola can be expressed in standard form as either (x - h)^2 = 4p(y - k) for vertical parabolas or (y - k)^2 = 4p(x - h) for horizontal parabolas. This form helps identify the vertex (h, k), the direction the parabola opens, and the distance p from the vertex to the focus and directrix.

Focus and Directrix of a Parabola

The focus is a fixed point inside the parabola, and the directrix is a line outside it, both equidistant from any point on the parabola. For the equation x^2 = -4y, the focus lies below the vertex, and the directrix is a horizontal line above it, determined by the value of p.

Graphing and Matching Parabolas

To match an equation to its graph, analyze the parabola's orientation, vertex, and key features like focus and directrix. Understanding how the sign and magnitude of coefficients affect the parabola's shape and position is essential for correctly identifying the corresponding graph.

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Horizontal Parabolas

Related Practice

Textbook Question

Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: (-4,0), (4,0); Vertices: (-5,0) (5,0)

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Graph each ellipse and locate the foci. x²/9 +y²/36= 1

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Graph the ellipse and locate the foci. (y^2)/25 + (x^2)/16 = 1

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Find the vertices and locate the foci of each hyperbola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d).

a. b. c. d.

y²/4−x²/1=1

Textbook Question

Graph the ellipse and locate the foci. 9x^2 + 4y^2 - 18x + 8y -23 = 0

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Textbook Question

In Exercises 1–4, find the focus and directrix of each parabola with the given equation. Then match each equation to one of the graphs that are shown and labeled (a)–(d). y^2 = - 4x