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Parabolas

SECTION 8.1

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Parabolas

The graph of a circle, an ellipse, a parabola, or a hyperbola can be formed by the intersection of a plane and a cone. Hence, these figures are referred to as conic sections. See Figure 8.1.

Figure 8.1

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Parabolas

A plane perpendicular to the axis of the cone intersects the cone in a circle (plane C). The plane E, tilted so that it is not perpendicular to the axis, intersects the cone in an ellipse.

When the plane is parallel to a line on the surface of the cone, the plane intersects the cone in a parabola. When the plane intersects both portions of the cone, a hyperbola is formed.

If the intersection of a plane and a cone is a point, a line, or two intersecting lines, then the intersection is called a degenerate conic section.

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Parabolas with Vertex at (0, 0)

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Parabolas with Vertex at (0, 0)

In addition to the geometric description of a conic section just given, a conic section can be defined as a set of points.

This method uses specified conditions about a curve to determine which points in the coordinate system are points of the graph. For example, a parabola can be defined by the following set of points.

Definition of a Parabola

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

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Parabolas with Vertex at (0, 0)

The line that passes through the focus and is perpendicular to the directrix is called the axis of symmetry of the parabola.

The midpoint of the line segment between the focus and the directrix on the axis of symmetry is the vertex of the parabola, as shown in Figure 8.2.

Figure 8.2

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Parabolas with Vertex at (0, 0)

Using this definition of a parabola, we can determine an equation of a parabola.

Suppose that the coordinates of the vertex of a parabola are V(0, 0) and the axis of symmetry is the y-axis. The equation of the directrix is y = –p, p > 0.

The focus lies on the axis of symmetry and is the same distance from the vertex as the vertex is from the directrix.

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Parabolas with Vertex at (0, 0)

Thus the coordinates of the focus are F(0, p), as shown in Figure 8.3.

Let P(x, y) be any point P on the parabola. Then, using the distance formula and the fact that the distance between any point P on the parabola and the focus is equal to the distance from the point P to the directrix, we can write the equation

d(P, F ) = d(P, D)

Figure 8.3

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Parabolas with Vertex at (0, 0)

By the distance formula,

Now, squaring each side and simplifying, we get

x2 + y2 – 2py + p2 = y2 + 2py + p2

x2 = 4py

This is the standard form of the equation of a parabola with vertex at the origin and the y-axis as its axis of symmetry.

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Parabolas with Vertex at (0, 0)

The standard form of the equation of a parabola with vertex at the origin and the x-axis as its axis of symmetry is derived in a similar manner.

Standard Forms of the Equation of a Parabola with Vertex at the Origin

Axis of Symmetry Is the y-Axis

The standard form of the equation of a parabola with vertex (0, 0) and the y-axis as its axis of symmetry is

x2 = 4py

The focus is (0, p), and the equation of the directrix is y = –p.

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Parabolas with Vertex at (0, 0)

If p > 0, the graph of the parabola opens up. See Figure 8.4a

Figure 8.4(a)

The graph of x2 = 4py with p > 0

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Parabolas with Vertex at (0, 0)

If p < 0, the graph of the parabola opens down. See Figure 8.4b.

Figure 8.4(b)

The graph of x2 = 4py with p < 0

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Parabolas with Vertex at (0, 0)

Axis of Symmetry Is the x-Axis

The standard form of the equation of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry is

y2 = 4px

The focus is (p, 0), and the equation of the directrix is x = –p.

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Parabolas with Vertex at (0, 0)

If p > 0, the graph of the parabola opens to the right.

See Figure 8.4c.

The graph of y2 = 4px with p > 0

Figure 8.4(c)

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Parabolas with Vertex at (0, 0)

If p < 0, the graph of the parabola opens to the left.

See Figure 8.4d.

The graph of y2 = 4px with p < 0

Figure 8.4(d)

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Example 1 – Find the Focus and Directrix of a Parabola

Find the focus and directrix of the parabola given by the

equation

Solution:

Because the x term is squared, the standard form of the equation is x2 = 4py.

x2 = –2y

Comparing this equation with x2 = 4py gives

4p = –2

Write the given equation in standard form.

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Example 1 – Solution

Because p is negative, the parabola opens down, and the focus is below the vertex (0, 0), as shown in Figure 8.5.

The coordinates of the focus are

The equation of the directrix is

Figure 8.5

cont’d

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Parabolas with Vertex at (h, k)

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Parabolas with Vertex at (h, k)

The equation of a parabola with a vertical or horizontal axis of symmetry and with vertex at a point (h, k) can be found by using the translations.

Consider a coordinate system with coordinate axes labeled x and y  placed so that its origin is at (h, k) of the xy-coordinate system.

The relationship between an ordered pair in the xy -coordinate system and one in the xy-coordinate system is given by the transformation equations

x = x – h

y  = y – k

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Parabolas with Vertex at (h, k)

Now consider a parabola with vertex at (h, k), as shown in Figure 8.7.

Create a new coordinate system with axes labeled x and y  and with its origin at (h, k).

The equation of a parabola in the xy -coordinate system is

(x)2 = 4py 

Figure 8.7

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Parabolas with Vertex at (h, k)

Using the transformation equations, we can substitute the expressions for x and y  into the equation above.

The standard form of the equation of a parabola with vertex (h, k) and a vertical axis of symmetry is

(x – h)2 = 4p( y – k)

Similarly, we can derive the standard form of the equation of a parabola with vertex (h, k) and a horizontal axis of symmetry.

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Parabolas with Vertex at (h, k)

Standard Forms of the Equation of a Parabola with Vertex at (h, k)

Vertical Axis of Symmetry

The standard form of the equation of a parabola with vertex (h, k) and a vertical axis of symmetry is

(x – h)2 = 4p( y – k)

The focus is (h, k + p), and the equation of the directrix is y = k – p.

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Parabolas with Vertex at (h, k)

If p > 0, the parabola opens up. See Figure 8.8a

Figure 8.8(a)

The graph of (x – h)2 = 4p(y – k) with p > 0

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Parabolas with Vertex at (h, k)

If p < 0, the parabola opens down. See Figure 8.8b

Figure 8.8(b)

The graph of (x – h)2 = 4p(y – k) with p < 0

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Parabolas with Vertex at (h, k)

Horizontal Axis of Symmetry

The standard form of the equation of a parabola with vertex (h, k) and a horizontal axis of symmetry is

(y – k)2 = 4p( x – h)

The focus is (h + p, k), and the equation of the directrix is x = h – p.

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Parabolas with Vertex at (h, k)

If p > 0, the parabola opens to the right. See Figure 8.8c.

The graph of (y – k)2 = 4p(x – h) with p > 0

Figure 8.8(c)

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Parabolas with Vertex at (h, k)

If p < 0, the parabola opens to the left. See Figure 8.8d.

In Example 3, we complete the square to find the standard form of a parabola and then use the standard form to determine the vertex, focus, and directrix of the parabola.

Figure 8.8(d)

The graph of (y – k)2 = 4p(x – h) with p < 0

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Example 3 – Find the Focus and Directrix of a Parabola

Find the equation of the directrix and the coordinates of the vertex and focus of the parabola given by the equation

3x + 2y2 + 8y – 4 = 0.

Solution:

Rewrite the equation so that the y terms are on one side of the equation, and then complete the square on y.

3x + 2y2 + 8y – 4 = 0

2y2 + 8y = –3x + 4

2( y2 + 4y) = –3x + 4

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Example 3 – Solution

2( y2 + 4y + 4) = –3x + 4 + 8

2( y + 2)2 = –3(x – 4)

Comparing this equation to (y – k)2 = 4p( x – h), we have a parabola that opens to the left with vertex (4, –2) and

Thus

Complete the square. Note that 2 • 4 = 8 is added to each side.

Simplify and factor.

Write the equation in standard form.

cont’d

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Example 3 – Solution

The coordinates of the focus are

The equation of the directrix is

cont’d

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Example 3 – Solution

Choosing some values for y and finding the corresponding values for x, we plot a few points.

Because the line y = –2 is the axis of symmetry, for each point on one side of the axis of symmetry there is a corresponding point on the other side.

Two points are (–2, 1) and (–2, –5). See Figure 8.9.

cont’d

Figure 8.9

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Ellipses

SECTION 8.2

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Ellipses

An ellipse is another of the conic sections formed when a plane intersects a right circular cone.

If  is the angle at which the plane intersects the axis of the cone and  is the angle shown in Figure 8.17, an ellipse is formed when  <  < 90.

If  = 90, then a circle is formed.

Figure 8.17

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Ellipses

As was the case for a parabola, there is a definition for an ellipse in terms of a certain set of points in the plane.

Definition of an Ellipse

An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, foci, is a positive constant. The midpoint of the line segment that connects the foci is the center of the ellipse.

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Ellipses

Equipped only with a piece of string and two tacks, we can use this definition to draw an ellipse (see Figure 8.18).

Tack the ends of the string to the foci, and trace a curve with a pencil held tight against the string.

The resulting curve is an ellipse.

The positive constant mentioned

in the definition of an ellipse is the

length of the string.

Figure 8.18

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Ellipses with Center at (0, 0)

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Ellipses with Center at (0, 0)

The graph of an ellipse has two axes of symmetry (see Figure 8.19).

The longer axis is called the major axis. The foci of the ellipse are on the major axis.

The shorter axis is called the minor axis.

It is customary to denote the length of the major axis by 2a and the length of the minor axis by 2b.

Figure 8.19

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Ellipses with Center at (0, 0)

The center of the ellipse is the midpoint of the major axis.

The endpoints of the major axis are the vertices (plural of vertex) of the ellipse.

A semimajor axis of an ellipse is a line segment that connects the center point of the ellipse with a vertex. Its length is half the length of the major axis.

A semiminor axis of an ellipse is a line segment that lies on the minor axis and connects the center point with a point on the ellipse. Its length is half the length of the minor axis.

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Ellipses with Center at (0, 0)

Consider the point V2(a, 0), which is one vertex of an ellipse, and the points F2(c, 0) and F1(–c, 0), which are the foci of the ellipse shown in Figure 8.20.

The distance from V2 to F1 is a + c. Similarly, the distance from V2 to F2 is a – c.

From the definition of an ellipse, the sum of the distances from any point on the ellipse to the foci is a positive constant.

Figure 8.20

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Ellipses with Center at (0, 0)

By adding a + c and a – c, we have

(a + c) + (a – c) = 2a

Thus the positive constant referred to in the definition of an ellipse is 2a, the length of the major axis.

Now let P(x, y) be any point on the ellipse (see Figure 8.21).

Figure 8.21

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Ellipses with Center at (0, 0)

By using the definition of an ellipse, we have

d(P, F1) + d(P, F2) = 2a

Subtract the second radical from each side of the equation, and then square each side of the equation.

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Ellipses with Center at (0, 0)

Divide by – 4, and

then square each side.

Rewrite with x and y

terms on the left side.

Factor.

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Ellipses with Center at (0, 0)

Divide each side by – a2b2.

The result is an equation of an

ellipse with center at (0, 0).

Let b2 = a2 – c2.

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Ellipses with Center at (0, 0)

Standard Forms of the Equation of an Ellipse with Center at the Origin

Major Axis on the x-Axis

The standard form of the equation of an ellipse with center at the origin and major axis on the x-axis (see Figure 8.22a) is given by

The length of the major axis is 2a.

The length of the minor axis is 2b.

Figure 8.22(a)

Major axis on the x-axis

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Ellipses with Center at (0, 0)

The coordinates of the vertices are (a, 0) and (–a, 0), and the coordinates of the foci are (c, 0) and (–c, 0), where c2 = a2 – b2.

Major Axis on the y-Axis

The standard form of the equation of an ellipse with center at the origin and major axis on the y-axis (see Figure 8.22b) is given by

Figure 8.22(b)

Major axis on the x-axis

Major axis on the y-axis

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Ellipses with Center at (0, 0)

The length of the major axis is 2a.The length of the minor

axis is 2b.

The coordinates of the vertices are (0, a) and (0, –a), and the coordinates of the foci are (0, c) and (0, –c), where c2 = a2 – b2.

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Example 1 – Find the Vertices and Foci of an Ellipse

Find the vertices and foci of the ellipse given by the

equation Sketch the graph.

Solution:

Because the y2 term has the larger denominator, the major axis is on the y-axis.

a2 = 49

a = 7

b2 = 25

b = 5

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Example 1 – Solution

c2 = a2 – b2

= 49 – 25

= 24

c =

The vertices are (0, 7) and (0, –7).

The foci are and

See Figure 8.23.

Figure 8.23

cont’d

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Ellipses with Center at (0, 0)

An ellipse with foci (3, 0) and (–3, 0) and major axis of length 10 is shown in Figure 8.24.

To find the equation of the ellipse in standard form, we must find a2 and b2.

Because the foci are on the major axis, the major axis is on the x-axis.

The length of the major axis is 2a.

Thus 2a = 10.

Figure 8.24

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Ellipses with Center at (0, 0)

Solving for a, we have a = 5 and a2 = 25.

Because the foci are (3, 0) and (–3, 0) and the center of the ellipse is the midpoint between the two foci, the distance from the center of the ellipse to a focus is 3.

Therefore, c = 3.

To find b2 use the equation

c2 = a2 – b2

9 = 25 – b2

b2 = 16

The equation of the ellipse in standard form is

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Ellipses with Center at (h, k)

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Ellipses with Center at (h, k)

The equation of an ellipse with center at (h, k) and with a

horizontal major axis can be found by using a translation of coordinates.

On a coordinate system with axes labeled x and y, the standard form of the equation of an ellipse with center at the origin of the xy-coordinate system is

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Ellipses with Center at (h, k)

Now place the origin of the xy-coordinate system at (h, k) in an xy-coordinate system. See Figure 8.25.

The relationship between an ordered pair in the xy-coordinate system and one in the xy-coordinate system is given by the transformation equations

x = x – h

y = y – k

Figure 8.25

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Ellipses with Center at (h, k)

Substitute the expressions for x and y into the equation of

an ellipse.

The equation of the ellipse with center at (h, k) is

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Ellipses with Center at (h, k)

Standard Forms of the Equation of an Ellipse with Center at (h, k)

Major Axis Parallel to the x-Axis

The standard form of the equation of an ellipse with center at (h, k) and major axis parallel to the x-axis (see Figure 8.26a) is given by

The length of the major axis is 2a.

The length of the minor axis is 2b.

Figure 8.26(a)

Major axis parallel to the x-axis

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Ellipses with Center at (h, k)

The coordinates of the vertices are (h + a, k) and (h – a, k), and the coordinates of the foci are (h + c, k) and (h – c, k), where c2 = a2 – b2.

Major Axis Parallel to the y-Axis

The standard form of the equation of an ellipse with center at (h, k) and major axis parallel to the y-axis (see Figure 8.26b) is given by

Figure 8.26(b)

Major axis parallel to the y-axis

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Ellipses with Center at (h, k)

The length of the major axis is 2a.

The length of the minor axis is 2b.

The coordinates of the vertices are (h, k + a) and (h, k – a), and the coordinates of the foci are (h, k + c) and (h, k – c), where c2 = a2 – b2.

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Example 2 – Find the Center, Vertices, and Foci of an Ellipse

Find the center, vertices, and foci of the ellipse given by the equation 4x2 + 9y2 – 8x + 36y + 4 = 0. Sketch the graph.

Solution:

Write the equation of the ellipse in standard form by completing the square.

4x2 + 9y2 – 8x + 36y + 4 = 0

4x2 – 8x + 9y2 + 36y = –4

4(x2 – 2x) + 9(y2 + 4y) = –4

Rearrange terms.

Factor.

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Example 2 – Solution

4(x2 – 2x + 1) + 9(y2 + 4y + 4) = –4 + 4 + 36

4(x – 1)2 + 9(y + 2)2 = 36

From the equation of the ellipse in standard form, the coordinates of the center of the ellipse are (1, –2).

Because the larger denominator is 9, the major axis is parallel to the x-axis and a2 = 9. Thus a = 3.

The vertices are (4, –2) and (–2, –2).

Complete the square.

Factor.

cont’d

Divide each side by 36.

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Example 2 – Solution

To find the coordinates of the foci, we find c.

c2 = a2 – b2

= 9 – 4

= 5

c =

The foci are and

See Figure 8.27.

cont’d

Figure 8.27

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Hyperbolas

SECTION 8.3

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Hyperbolas

A hyperbola is a conic section formed when a plane intersects a right circular cone at a certain angle.

If  is the angle at which the

plane intersects the axis of

the cone and  is the angle

shown in Figure 8.35, a

hyperbola is formed when

0° <  <  or when the plane

is parallel to the axis of the

cone.

Figure 8.35

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Hyperbolas

As with the other conic sections, there is a definition of a hyperbola in terms of a certain set of points in the plane.

Definition of a Hyperbola

A hyperbola is the set of all points in the plane the difference between whose distances from two fixed points, foci, is a positive constant.

This definition differs from that of an ellipse in that the ellipse was defined in terms of the sum of two distances, whereas the hyperbola is defined in terms of the difference of two distances.

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Hyperbolas with Center at (0, 0)

The transverse axis of a hyperbola shown in Figure 8.36 is the line segment joining the intercepts.

The midpoint of the transverse

axis is called the center of the

hyperbola.

The conjugate axis is a line

segment that passes through

the center of the hyperbola

and is perpendicular to the

transverse axis.

Figure 8.36

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Hyperbolas with Center at (0, 0)

The length of the transverse axis is customarily

represented as 2a, and the distance between the two foci is

represented as 2c. The length of the conjugate axis is

represented as 2b.

The vertices of a hyperbola are the points where the

hyperbola intersects the transverse axis.

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Hyperbolas with Center at (0, 0)

To determine the positive constant stated in the definition

of a hyperbola, consider the point V1(a, 0), which is one

vertex of a hyperbola, and the points F1(c, 0) and F2(–c, 0),

which are the foci of the hyperbola (see Figure 8.37).

Figure 8.37

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Hyperbolas with Center at (0, 0)

The difference between the distance from V1(a, 0) to

F1(c, 0), c – a, and the distance from V1(a, 0) to

F2(–c, 0), c + a, must be a constant.

By subtracting these distances, we find

| (c – a) – (c + a) | = | –2a | = 2a

Thus the constant is 2a, and it is the length of the

transverse axis. The absolute value is used to ensure that

the distance is a positive number.

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Hyperbolas with Center at (0, 0)

Standard Forms of the Equation of a Hyperbola with Center at the Origin

Transverse Axis on the x-Axis

The standard form of the equation

of a hyperbola with center at the

origin and transverse axis on the

x-axis (see Figure 8.38a) is given

Figure 8.38(a)

Transverse axis on the x-axis

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Hyperbolas with Center at (0, 0)

The coordinates of the vertices are (a, 0) and (–a, 0), and

the coordinates of the foci are (c, 0) and (–c, 0), where

c2 = a2 + b2.

Transverse Axis on the y-Axis

The standard form of the equation

of a hyperbola with center at the

origin and transverse axis on the

y-axis (see Figure 8.38b) is given

Figure 8.38(b)

Transverse axis on the y-axis

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Hyperbolas with Center at (0, 0)

The coordinates of the vertices are (0, a) and (0, –a), and the coordinates of the foci are (0, c) and (0, – c), where c2 = a2 + b2.

By looking at the equations, it is possible to determine the

location of the transverse axis by finding which term in the

equation is positive.

When the x2 term is positive, the transverse axis is on the

x-axis. When the y2 term is positive, the transverse axis is

on the y-axis.

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Hyperbolas with Center at (0, 0)

Consider the hyperbola given by the equation

Because the x2 term is positive, the transverse axis is on

the x-axis, a2 = 16, and thus a = 4.

The vertices are (4, 0) and (–4, 0). To find the foci, we

determine c.

c2 = a2 + b2 = 16 + 9 = 25

c = = 5

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Hyperbolas with Center at (0, 0)

The foci are (5, 0) and (–5, 0). The graph is shown in

Figure 8.39.

Each hyperbola has two asymptotes that pass through the

center of the hyperbola. The asymptotes of the hyperbola

are a useful guide to sketching the graph of the hyperbola.

Figure 8.39

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Hyperbolas with Center at (0, 0)

Asymptotes of a Hyperbola with Center at the Origin

The asymptotes of the hyperbola defined by

are given by the equations and

(see Figure 8.40a).

Figure 8.40(a)

Asymptotes of

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Hyperbolas with Center at (0, 0)

The asymptotes of the hyperbola defined by

are given by the equations and

(see Figure 8.40b).

Figure 8.40(b)

Asymptotes of

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Hyperbolas with Center at (0, 0)

One method for remembering the equations of the

asymptotes is to write the equation of a hyperbola in

standard form and then replace 1 with 0 and solve for y.

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Example 1 – Find the Vertices, Foci, and Asymptotes of a Hyperbola

Find the vertices, foci, and asymptotes of the hyperbola

given by the equation Sketch the graph.

Solution:

Because the y2 term is positive, the transverse axis is on the y-axis. We know that a2 = 9; thus a = 3.

The vertices are V1(0, 3) and V2(0, –3).

c2 = a2 + b2

= 9 + 4

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Example 1 – Solution

The foci are F1(0, ) and F2(0, ).

Because a = 3 and b = 2 (b2 = 4), the equations of the

asymptotes are and

To sketch the graph, we draw a rectangle that has its

center at the origin and has dimensions equal to the

lengths of the transverse and conjugate axes.

cont’d

‹#›

Example 1 – Solution

The asymptotes are extensions of the diagonals of the

rectangle. See Figure 8.41.

Figure 8.41

cont’d

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Hyperbolas with Center at (h, k)

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Hyperbolas with Center at (h, k)

Using a translation of coordinates similar to that used for

ellipses, we can write the equation of a hyperbola with

center at the point (h, k).

Given coordinate axes labeled x  and y , an equation of a

hyperbola with center at the origin is

(1)

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Hyperbolas with Center at (h, k)

Now place the origin of this coordinate system at the point

(h, k) of the xy-coordinate system, as shown in Figure 8.42.

The relationship between an ordered

pair in the x y -coordinate system and

one in the xy-coordinate system is

given by the transformation equations

x  = x – h

y  = y – k

Figure 8.42

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Hyperbolas with Center at (h, k)

Substitute the expressions for x  and y  into Equation (1).

The equation of a hyperbola with center at (h, k) is

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Hyperbolas with Center at (h, k)

Standard Forms of the Equation of a Hyperbola with Center at (h, k)

Transverse Axis Parallel to the x-Axis

The standard form of the equation

of a hyperbola with center at (h, k)

and transverse axis parallel to the

x-axis (see Figure 8.43a) is given

Transverse axis parallel to the x-axis

Figure 8.43(a)

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Hyperbolas with Center at (h, k)

The coordinates of the vertices are V1(h + a, k) and V2(h – a, k).

The coordinates of the foci are F1(h + c, k) and F2(h – c, k),

where c2 = a2 + b2.

The equations of the asymptotes are

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Hyperbolas with Center at (h, k)

Transverse Axis Parallel to the y-Axis

The standard form of the equation of a hyperbola with

center at (h, k) and transverse axis parallel to the y-axis

(see Figure 8.43b) is given by

Figure 8.43(b)

Transverse axis parallel to the y-axis

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Hyperbolas with Center at (h, k)

The coordinates of the vertices are V1(h, k + a) and V2(h, k – a).

The coordinates of the foci are F1(h, k + c) and F2(h, k – c),

where c2 = a2 + b2.

The equations of the asymptotes are

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Example 2 – Find the Center, Vertices, Foci, and Asymptotes of a Hyperbola

Find the center, vertices, foci, and asymptotes of the hyperbola given by the equation 4x2 – 9y2 – 16x + 54y – 29 = 0. Sketch the graph.

Solution:

Write the equation of the hyperbola in standard form by completing the square.

4x2 – 9y2 – 16x + 54y – 29 = 0

4x2 – 16x – 9y2 + 54y = 29

4(x2 – 4x) – 9(y2 – 6y) = 29

4(x2 – 4x + 4) – 9(y2 – 6y + 9) = 29 + 16 – 81

Rearrange terms.

Factor.

Complete the

square.

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Example 2 – Solution

4(x – 2)2 – 9(y – 3)2 = –36

The coordinates of the center are (2, 3).

Because the term containing (y – 3)2 is positive, the

transverse axis is parallel to the y-axis. We know that

a2 = 4; thus a = 2.

Divide each side by –36.

Factor.

cont’d

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Example 2 – Solution

The vertices are (2, 5) and (2, 1). See Figure 8.44.

To find the coordinates of the

foci, we find c.

c2 = a2 + b2

= 4 + 9

The foci are (2, 3 + ) and

(2, 3 – ).

Figure 8.44

cont’d

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Example 2 – Solution

We know that b2 = 9; thus b = 3.

The equations of the asymptotes are ,

which simplifies to

cont’d

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