College AlgebraFifth EditionJames Stewart Lothar Redlin Saleem Watson

Conic Sections9

Chapter Overview

Conic sections are the curves we get when we make a straight cut in a cone.

• For example, if a cone is cut horizontally, the cross section is a circle.

• So, a circle is a conic section.

Chapter Overview

Other ways of cutting a cone produce parabolas, ellipses, and hyperbolas.

Chapter Overview

Our goal in this chapter is to find equations whose graphs are the conic sections.

• We already know from Section 2.2 that the graph of the equation x2 + y2 = r2 is a circle.

• We will find equations for each of the other conic sections by analyzing their geometric properties.

Parabolas8.1

Parabolas

We saw in Section 4.1 that the graph of the equation y = ax2 + bx + c is a U-shaped curve called a parabola that opens either upward or downward—depending on whether the sign of a is positive or negative.

• Here, we study parabolas from a geometric rather than an algebraic point of view.

Parabolas

We begin with the geometric definition of a parabola and show how this leads to the algebraic formula that we are already familiar with.

Parabola—Geometric Definition

A parabola is the set of points in the plane equidistant from a fixed point F (focus) and a fixed line l (directrix).

• The vertex V lies halfway between the focus and the directrix.

• The axis of symmetry is the line that runs through the focus perpendicular to the directrix.

Parabolas

In this section, we restrict our attention to parabolas that:

• Are situated with the vertex at the origin.• Have a vertical or horizontal axis of symmetry.

Parabolas in more general positions will be considered in Section 8.4.

Parabolas

If the focus of such a parabola is the point F(0, p), the axis of symmetry must be vertical and the directrix has the equation y = –p.

• The figure illustrates the case p > 0.

Parabolas

If P(x, y) is any point on the parabola,

• The distance from P to the focus F (using the Distance Formula) is:

• The distance from P to the directrix is: | y – (–p) | = | y + p |

2 2( )x y p

Parabolas

By the definition of a parabola, these two distances must be equal:

2 2

22 2 2

2 2 2 2 2

2

2

( )

( ) ( )

2 2

2 2

4

x y p y p

x y p y p y p

x y py p y py px py py

x py

Parabolas

If p > 0, then the parabola opens upward.

If p < 0, it opens downward.

When x is replaced by –x, the equation remains unchanged.

• So, the graph is symmetric about the y-axis.

Equations and Graphs of Parabola

Equations and Graphs of Parabola

We now summarize what we have just proved about the equation and features of a parabola with a vertical axis.

Parabola with Vertical Axis

The graph of the equation x2 = 4py

is a parabola with these properties.

Vertex V(0, 0)

Focus F(0, p)

Directrix y = –p

Parabola with Vertical Axis

The parabola opens:• Upward if p > 0.• Downward if p < 0.

E.g. 1—Finding the Equation of a Parabola

Find the equation of the parabola with vertex V(0, 0) and focus F(0, 2), and sketch its graph.

• Since the focus is F(0, 2), we conclude that p = 2 (and so the directrix is y = –2).

• Thus, the equation is: x2 = 4(2)y x2 = 8y

E.g. 1—Finding the Equation of a Parabola

Since p = 2 > 0, the parabola opens upward.

E.g. 2—Finding Focus and Directrix from Equation

Find the focus and directrix of the parabola y = –x2, and sketch the graph.

• To find the focus and directrix, we put the given equation in the standard form x2 = –y.

• Comparing this to the general equation x2 = 4py, we see that 4p = –1; so, p = –¼.

• The focus is F(0, –¼) and the directrix is y = ¼.

E.g. 2—Finding Focus and Directrix from Equation

Here’s the graph of the parabola, together with the focus and the directrix.

E.g. 2—Finding Focus and Directrix from Equation

We can also draw the graph using a graphing calculator.

Equations and Graphs of Parabola

Reflecting the graph in this figure about the diagonal line y = x has the effect of interchanging the roles of x and y.

• This results in a parabola with horizontal axis.

• By the same method as before, we can prove the following properties.

Parabola with Horizontal Axis

The graph of the equation y2 = 4px

is a parabola with these properties.

Vertex V(0, 0)

Focus F(p, 0)

Directrix x = –p

Parabola with Horizontal Axis

The parabola opens:• To the right if p > 0.• To the left if p < 0.

E.g. 3—Parabola with Horizontal Axis

A parabola has the equation 6x + y2 = 0

(a) Find the focus and directrix of the parabola, and sketch the graph.

(b) Use a graphing calculator to draw the graph.

E.g. 3—Parabola with Horiz. Axis

We put the given equation in the standard form y2 = –6x.

• Comparing this to the general equation y2 = 4px, we see that 4p = –6; so, p = (–3/2).

• The focus is F(–3/2, 0) and the directrix is x = 3/2.

Example (a)

E.g. 3—Parabola with Horiz. Axis

Since p < 0, the parabola opens to the left.

Example (a)

E.g. 3—Parabola with Horiz. Axis

To draw the graph using a graphing calculator, we need to solve for y.

6x + y2 = 0

y2 = –6x6y x

Example (b)

E.g. 3—Parabola with Horiz. Axis

To obtain the graph of the parabola, we graph both functions.

Example (b)

Note

The equation y2 = 4px does not define y as a function of x.

• So, to use a graphing calculator to graph a parabola with horizontal axis, we must first solve for y.

• This leads to two functions

• We need to graph both to get the complete graph of the parabola.

4 and 4

6 and 6

y px y px

y x y x

Width of Parabola

We can use the coordinates of the focus to estimate the “width” of a parabola when sketching its graph.

• The line segment that runs through the focus perpendicular to the axis—with endpoints on the parabola—is called the latus rectum.

• Its length is the focal diameter of the parabola.

Width of Parabola

From the figure, we can see that the distance from an endpoint Q of the latus rectum to the directrix is | 2p |.

• So, the distance from Q to the focus must be | 2p | as well (by the definition of a parabola).

• Hence, the focal diameter is | 4p |.

Width of Parabola

In the next example, we use the focal diameter to determine the “width” of a parabola when graphing it.

E.g. 4—Focal Diameter of a Parabola

Find the focus, directrix, and focal diameter of the parabola y = ½x2, and sketch its graph.

• We put the equation in the form x2 = 4py.y = ½x2

x2 = 2y• We see that 4p = 2.• So, the focal diameter is 2.

E.g. 4—Focal Diameter of a Parabola

Solving for p gives p = ½.

• So, the focus is (0, ½) and the directrix is y = –½ .

Since the focal diameter is 2, the latus rectum extends 1 unit to the left and 1 unit to the right

of the focus.

E.g. 4—Focal Diameter of a Parabola

Here’s the graph.

Family of Parabolas

In the next example, we graph a family of parabolas—to show how changing the distance between the focus and the vertex affects the “width” of a parabola.

E.g. 5—Family of Parabolas

(a) Find equations for the parabolas with vertex at the origin and foci

(b) Draw the graphs of the parabolas in (a).

• What do you conclude?

1 11 2 3 48 20, , 0, , 0, 1 , (0, 4)F F F F

E.g. 5—Family of Parabolas

Since the foci are on the positive y-axis, the parabolas open upward and have equations of the form x2 = 4py.

• This leads to the following equations.

Example (a)

E.g. 5—Family of Parabolas

We see that the closer the focus to the vertex, the narrower the parabola.

Example (b)

Applications

Applications

Parabolas have an important property that makes them useful as reflectors for lamps and telescopes.

Applications

Light from a source placed at the focus of a surface with parabolic cross section will be reflected in such a way that it travels parallel to the axis of the parabola.

• Thus, a parabolic mirror reflects the light into a beam of parallel rays.

Reflection Property

Conversely, light approaching the reflector in rays parallel to its axis of symmetry is concentrated to the focus.

• This reflection property—which can be proved using calculus—is used in the construction of reflecting telescopes.

E.g. 6—Focal Point of a Searchlight Reflector

A searchlight has a parabolic reflector that forms a “bowl,” 12 in. wide from rim to rim and 8 in. deep.

• If the filament of the light bulb is located at the focus, how far from the vertex of the reflector is it?

E.g. 6—Focal Point of a Searchlight Reflector

We introduce a coordinate system and place a parabolic cross section of the reflector so that:

• Its vertex is at the origin.• Its axis is vertical.

E.g. 6—Focal Point of a Searchlight Reflector

Then, the equation of this parabola has the form x2 = 4py.

• We see that the point (6, 8) lies on the parabola.

• We use this to find p.

E.g. 6—Focal Point of a Searchlight Reflector

62 = 4p(8) 36 = 32p p = (9/8)

• The focus is F(0, (9/8)).• So, the distance between the vertex

and the focus is in.• Since the filament is positioned at the focus,

it is located in. from the vertex of the reflector.

9 18 81

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