[PDF] 9.3 The Parabola Write as a quadratic equation

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900Chapter 9

79.Write as a quadratic

equation in and then use the quadratic formula to express in terms of Graph the resulting two equations using a graphing utility in a by viewing rectangle.What effect does the have on the graph of the resulting hyperbola? What problems would you encounter if you attempted to write the given equation in standard form by completing the square?

80.Graph and in the same viewing

rectangle.Explain why the graphs are not the same.

Critical Thinking Exercises

Make Sense?In Exercises 81Ð84, determine whether each statement makes sense or does not make sense, and explain your reasoning.

81.I changed the addition in an ellipse's equation to subtraction

and this changed its elongation from horizontal to vertical.

82.I noticed that the definition of a hyperbola closely resemblesthat of an ellipse in that it depends on the distances betweena set of points in a plane to two fixed points,the foci.

83.I graphed a hyperbola centered at the origin that had

but no

84.I graphed a hyperbola centered at the origin that wassymmetric with respect to the and also symmetricwith respect to the y-axis.x-axisx-intercepts.y-intercepts,x

†x†16-y†y†9=1x 2 16-y 2

9=1xy-term

3-30, 50, 1043-50, 70, 104x.yy4x

2 -6xy+2y2

-3x+10y-6=0In Exercises 85Ð88,determine whether each statement is true or false.If the statement is false, make the necessary change(s) to produce atrue statement.

85.If one branch of a hyperbola is removed from a graph, then

the branch that remains must define as a function of

86.All points on the asymptotes of a hyperbola also satisfy thehyperbola's equation.

87.The graph of does not intersect the line

88.Two different hyperbolas can never share the same asymptotes.

89.What happens to the shape of the graph of as

90.Find the standard form of the equation of the hyperbola withvertices and (5,6),passing through (0,9).

91.Find the equation of a hyperbola whose asymptotes areperpendicular.

Preview Exercises

Exercises 92Ð94 will help you prepare for the material covered in the next section. In Exercises 92Ð93, graph each parabola with the given equation.

92. 93.

94.Isolate the terms involving on the left side of the equation:

Then write the equation in an equivalent form by completing the square on the left side.y2 +2y+12x-23=0.yy=-31x-12 2 +2y=x 2 +4x-515, -62c a:q, where c 2 =a 2 +b 2 ?x 2a 2 -y 2 b 2 =1y=- 2 3 x.x 2 9-y 2

4=1x.y

9.3The Parabola

his NASA photograph is one of a series of stunning images captured from the ends of the universe by the Hubble Space

Telescope. The image shows infant

star systems the size of our solar system emerging from the gas and dust that shrouded their creation.

Using a parabolic mirror that is 94.5

inches in diameter, the Hubble has provided answers to many of the profound mysteries of the cosmos:

How big and how old is the

universe? How did the galaxies come to exist? Do other Earth-like planets orbit other sun-like stars? I n this section,we study parabolas and their applications, including parabolic shapes that gather distant rays of light and focus them into spectacular images.Definition of a Parabola In Chapter 2,we studied parabolas,viewing them as graphs of quadratic functions in the form y=a1x-h2 2 +kory=ax 2 +bx+c.

Graph parabolas with vertices

at the origin.

Write equations of parabolas

in standard form.Graph parabolas with verticesnot at the origin.

Solve applied problemsinvolving parabolas.

At first glance,this image looks like columns of smoke rising from a fire into a starry sky.Those are,indeed,stars in the background,but you are not looking at ordinary smoke columns.These stand almost

6 trillion miles high and are 7000 light-years from Earth - more

than 400 million times as far away as the sun. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 900

Section 9.3901

Study Tip

Here is a summary of what you should already know about graphing parabol as.

Graphing and

1.If the graph opens upward.If the graph opens downward.

2.The vertex of is

3.The of the vertex of is x=-

b 2 a.y=ax 2 +bx+cx-coordinate y y a x h 2 k a 0 xx y h k h k x h x h y a x h 2 k a 0 1 h k 2 .y=a1x-h2 2 +ka60,a70,yax 2 bxcya(xh) 2 k Parabolas can be given a geometric definition that enables us to include graphs that open to the left or to the right, as well as those that open obliquely.The definitions of ellipses and hyperbolas involved two fixed points, the foci. By contrast,the definition of a parabola is based on one point and a line. A parabolais the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus, that is not on the line (seeFigure 9.29).

Vertex

Directrix

Parabola

Axis of

symmetry Focus In Figure 9.29,find the line passing through the focus and perpendicular to the directrix. This is the axis of symmetryof the parabola. The point of intersection of the parabola with its axis of symmetry is called the vertex . Notice that the vertex is midway between the focus and the directrix.

Standard Form of the Equation of a Parabola

The rectangular coordinate system enables us

to translate a parabola's geometric definition into an algebraic equation.Figure 9.30is our starting point for obtaining an equation. We place the focus on the at the point

The directrix has an equation given by

The vertex,located midway between

the focus and the directrix,is at the origin.

What does the definition of a parabola

tell us about the point in

Figure 9.30

For any point on the parabola, the

distance to the directrix is equal to the distance to the focus.Thus,the point is on the parabola if and only if

Use the distance formula.

Square both sides of the

equation. 1x+p2 2 =1x-p2 2 +y 2

41x+p2

2 +1y-y2 2 =41x-p2 2 +1y-02 2 d 1 =d 2 .1x, y2d 2 d 1 1 x y

21x, y2x=-p.1p, 02.x-axis

Directrix:

x p

Focus (

p 0 x y

M(p, y)P(x, y)d

1 d 2 P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 901

902Chapter 9

Square and

Subtract from both

sides of the equation.

Solve for

This last equation is called the

standard form of the equation of a parabola with its vertex at the origin.There are two such equations, one for a focus on the and one for a focus on the y-axis.x-axis y 2 .y 2 =4px x 2 +p 2 2 px=-2px+y 2 x-p.x+px 2 +2px+p 2 =x 2 -2px+p 2 +y 2 The standard form of the equation of a parabolawith vertex at the origin is Figure 9.31(a)illustrates that for the equation on the left, the focus is on the which is the axis of symmetry.Figure 9.31(b)illustrates that for the equation on the right,the focus is on the which is the axis of symmetry.y-axis,x-axis,y 2 =4px or x 2 =4py.

Study Tip

It is helpful to think of as the

directed distancefrom the vertex to the focus. If the focus lies units to the right of the vertex or units above the vertex. If the focus lies units to the left of the p†p60,ppp70,p Using the Standard Form of the Equation of a Parabola We can use the standard form of the equation of a parabola to find its fo cus and directrix. Observing the graph's symmetry from its equation is helpful in locating the focus. Although the definition of a parabola is given in terms of its focus and its directrix,the focus and directrix are not part of the graph.The vertex,located at the origin,is a point on the graph of and Example 1 illustrates how you can find two additional points on the parabola.

Finding the Focus and Directrix of a Parabola

Find the focus and directrix of the parabola given by Then graph the parabola.y 2 =12x.

EXAMPLE 1

x 2 =4py.y 2 =4px y 2 =4pxx 2 =4py

The equation does not change if

y is replaced with y . There is x -axis symmetry and the focus is on the x -axis at ( p , 0).The equation does not change if x is replaced with x . There is y -axis symmetry and the focus is on the y -axis at (0, p

Graph parabolas with vertices at

the origin.

Directrix:

x p

Focus (

p , 0)Vertex x y y 2 4 px

Parabola with the as the

axis of symmetry.If the graph opens to the right.If the graph opens to the left.p60,p70,x-axis

Directrix:

y p x y x 2 4 py

Focus (0,

p

Vertex

Parabola with the

as the axis of symmetry.If the graph opens upward.If the graph opens downward.p60,p70,y-axis P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 902

Section 9.3903

The given equation,is in the standard form

so We can find both the focus and the directrix by finding

Divide both sides by 4.

Because is positive,the parabola,with its symmetry,opens to the right.Thequotesdbs_dbs12.pdfusesText_18

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