§ 10.4 the parabola; identifying conic sections. blitzer, intermediate algebra, 5e – slide #2...

§ 10.4

The Parabola; Identifying Conic Sections

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 10.4

Equation of a Parabola

We looked at parabolas in Chapter 8, viewing them as graphs of quadratic functions. In this section, we return again to the study of parabolas. Within this chapter on Conics, we extendour study by considering the formal definition of the parabola.

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

Note that the definition of both the hyperbola and the ellipse involved two fixed points, the foci. By contrast, the definition of a parabola is based on one point and a line.

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 10.4

Equation of a Parabola

Definition of a ParabolaA parabola is 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.

Directrix

FocusParabola Vertex

Axis of Symmetry

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 10.4

Equation of a Parabola

Parabolas Opening to the Left or to the RightThe graphs of are parabolas opening to the left or to the right.

1) If a > 0, the graph opens to the right. If a < 0, the graph opens to the left.

2) The vertex of

3) The y-coordinate of the vertex of

cbyayxhkyax 22 and

. is 2 h, khkyax

.2

is 2

a

bycbyayx

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 10.4

Equation of a Parabola

CONTINUECONTINUEDD

(h, k) y = k

0

2

a

hkyax

x

y

(h, k)y = k

0

2

a

hkyax

x

y

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 10.4

Equation of a Parabola

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Graph:

We can graph this equation by following the steps in the preceding box. We begin by identifying values for a, k, and h.

.12 2 yx

1) Determine how the parabola opens. Note that a, the coefficient of , is 1. Thus, a > 0; this positive value tells us that the parabola opens to the right.

12 2 yx

hkyax 2

2y

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 10.4

Equation of a Parabola

Graphing Horizontal ParabolasTo graph

1) Determine whether the parabola opens to the left or to the right. If a > 0, it opens to the right. If a < 0, it opens to the left.

2) Determine the vertex of the parabola. The vertex of is

(h, k). The y-coordinate of the vertex of

Substitute this value of y- into the equation to find the x-coordinate.

3) Find the x-intercept by replacing y with 0.

4) Find any y-intercepts by replacing x with 0. Solve the resulting quadratic equation for y.

5) Plot the intercepts and the vertex. Connect a more accurate graph, select values for y on each side of the axis of symmetry and compute values for x.

,or 22 cbyayxhkyax

hkyax 2

.2

is 2

a

bycbyayx

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 10.4

Equation of a Parabola

2) Find the vertex. The vertex of the parabola is at (h, k). Because k = 2 and h = 1, the parabola has its vertex at (1, 2).

3) Find the x-intercept. Replace y with 0 in

51412120 22 x

CONTINUECONTINUEDD

.12 2 yx

The x-intercept is 5. The parabola passes through (5, 0).

4) Find the y-intercepts. Replace x with 0 in the given equation. 12 2 yx This is the given equation.

120 2 y Replace x with 0.

221 y Subtract 1 from both sides.This equation clearly has no solutions since the left side is a negative number. Therefore, there are no y-intercepts.

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 10.4

Equation of a Parabola

CONTINUECONTINUEDD

5) Graph the parabola. With a vertex at (1, 2), an x-intercept at 5, and no y-intercepts, the graph of the parabola is shown as follows. The axis of symmetry is the horizontal line whose equation is y = 2.

-2

-1

0

1

2

3

4

5

6

0 2 4 6 8 10 12

Vertex: (1, 2)

x-intercept: (5, 0)

Axis of symmetry: y = 2.

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 10.4

Equation of a Parabola

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Graph:

1) Determine how the parabola opens. Note that a, the coefficient of , is -2. Thus a < 0; this negative value tells us that the parabola opens to the left.

.342 2 yyx

2) Find the vertex. We know that the y-coordinate of the vertex

is . We identify a, b, and c ina

by

2

2y

a = -2

342 2 yyx

b = 4 c = -3

.2 cbyayx

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 10.4

Equation of a Parabola

Substitute the values of a and b into the equation for the y-coordinate:

The y-coordinate of the vertex is 1. We substitute 1 for y into the parabola’s equation, to find the x-coordinate:

.122

4

2

a

by

342 2 yyx

CONTINUECONTINUEDD

.134231412 2 x

The vertex is at (-1, 1).

3) Find the x-intercept. Replace y with 0 in .342 2 yyx

330402 2 x

The x-intercept is -3. The parabola passes through (-3, 0).

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 10.4

Equation of a Parabola

CONTINUECONTINUEDD

4) Find the y-intercepts. Replace x with 0 in the given equation.

342 2 yyx This is the given equation.3420 2 yy Replace x with 0.

22

32444 2

yUse the quadratic formula to solve for y.

4

24164

y Simplify.

4

84

y Subtract.

This equation clearly has no solutions since the radicand is a negative number. Therefore, there are no y-intercepts.

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 10.4

Equation of a Parabola

CONTINUECONTINUEDD

5) Graph the parabola. With a vertex at (-1, 1), an x-intercept at -3, and no y-intercepts, the graph of the parabola is shown below. The axis of symmetry is the horizontal line whose equation is y = 1.

-3

-2

-1

0

1

2

3

4

5

-25 -20 -15 -10 -5 0

Vertex: (-1, 1)

x-intercept: (-3, 0)

Axis of symmetry: y = 1.

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 10.4

Equations of Conic Sections

Recognizing Conic Sections from EquationsConic Section How to Identify the Equation Example

Circle When - and -terms are on the same side, they have the same coefficient.

Ellipse When - and -terms are on the same side, they have different coefficients of the same sign.

or (dividing by 64)

Hyperbola When - and -terms are on the same side, they have coefficients with opposite signs.

or (dividing by 36)

Parabola Only one of the variables is squared.

2y2x

2y2x

2y2x

1622 yx

64164 22 yx

1416

22

yx

3649 22 xy

194

22

xy

542 yyx

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 10.4

Equations of Conic Sections

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola:

(Throughout the solution, in addition to identifying each equation’s graph, we’ll also discuss the graph’s important features.) If both variables are squared, the graph of the equation is not a parabola. In both cases, we collect the - and -terms on the same side of the equation.

.3273(b)436(a) 2222 yxyx

2y2x

22 436(a) yx The graph cannot be a parabola. To see if it is a circle, an ellipse, or a hyperbola, we collect the - and -terms on the same side. Add to both sides. We obtain the following:

2y2x24y

Blitzer, Intermediate Algebra, 5e – Slide #16 Section 10.4

Equations of Conic Sections

.364 22 yx

Because the - and -terms have different coefficients of the same sign, the equation’s graph is an ellipse.

2y2x

CONTINUECONTINUEDD

22 3273(b) yx

The graph cannot be a parabola. To see if it is a circle, an ellipse, or a hyperbola, we collect the - and -terms on the same side. Subtract from both sides. We obtain:

2y2x23y

2733 22 yx

Because the - and -terms have coefficients with opposite signs, the equation’s graph is a hyperbola.

2y2x

Blitzer, Intermediate Algebra, 5e – Slide #17 Section 10.4

The Parabola

A parabola is the set of all points that are equidistant from a fixed line, the directrix, and a fixed point, the focus, that is not on the line.

The line passing through the focus and perpendicular to the directrix is the axis of symmetry. The point of intersection of the parabola with its axis of symmetry is the vertex.