4.3. Conics

Objectives

• Recognize the four basic conics: circle, ellipse, parabola, and hyperbola.

• Recognize, graph, and write equations of parabolas (vertex at origin).

• Recognize, graph, and write equations of ellipses (center at origin).

• Recognize, graph, and write equations of hyperbolas (center at origin).

Introduction

Conic sections were discovered during the classical Greek period, 600 to 300 B.C.

This early Greek study was largely concerned with the geometric properties of conics.

It was not until the early 17th century that the broad applicability of conics became apparent and played a prominent role in the early development of calculus.

Introduction

A conic section (or simply conic) is the intersection of a plane and a double napped cone. Notice in Figure 4.16 that in the formation of the four basic conics, the intersecting plane does not pass through the vertex of the cone.

Figure 4.16

Basic Conics

Circle Ellipse Parabola Hyperbola

Introduction

When the plane does pass through the vertex, the resulting figure is a degenerate conic, as shown in Figure 4.17.

Figure 4.17

Degenerate Conics

Point Line Two Intersecting Lines

Introduction

There are several ways to approach the study of conics. You could begin by defining conics in terms of the intersections of planes and cones, as the Greeks did, or you could define them algebraically, in terms of the general second-degree equation

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.

However, you will study a third approach, in which each of the conics is defined as a locus (collection) of points satisfying a certain geometric property.

Introduction

For example, the definition of a circle as the collection of all points (x, y) that are equidistant from a fixed point (h, k) led easily to the standard form of the equation of a circle

(x – h)2 + (y – k)2 = r2.

We know that the center of a circle is at (h, k) and that the radius of the circle is r.

Equation of a circle

Circle

standard form of the equation of a circle

Circle with center at origin

x2 + y2 = r 2.

Example:

Find the equation of a circle whose center is

at the origin and which passes (2,0).

Answer

x2 + y2 = 2 2 because the radius is 2.

Example

1. Find the distance between the point (3,4) and (-1,2).

2. Find the equation of a circle which passes (3,4) and

whose center is at (-1,2).

Answer

1. Distance 𝑑 between 𝑥1, 𝑦1 𝑎𝑛𝑑 𝑥2, 𝑦2 = 𝑥1 − 𝑥22 + 𝑦1 − 𝑦2

2

Thus,

𝑑 = 3 − −12+ 4 − 2 2 = 3 + 1 2 + 22 = 42 + 4 = 16 + 4 = 20

2. (x + 1)2 + (y – 2)2 = 20.

Ellipses

Ellipses

The line through the foci intersects the ellipse at two points (vertices).

The chord joining the vertices is the

major axis, and its midpoint is the

center of the ellipse. The chord

perpendicular to the major axis at

the center is the minor axis.

(See Figure 4.21.)

Figure 4.21

Ellipses

You can visualize the definition of an ellipse by imagining two thumbtacks

placed at the foci, as shown in Figure 4.22.

If the ends of a fixed length of string

are fastened to the thumbtacks and

the string is drawn taut with a pencil,

the path traced by the pencil will be

an ellipse.

The standard form of the equation of an ellipse takes one of two forms,

depending on whether the major axis is horizontal or vertical.

Figure 4.22

Ellipses

(a) Major axis is horizontal;

minor axis is vertical.(b) Major axis is vertical;

minor axis is horizontal.

EllipsesIn Figure (a), note that because the sum of the distances from a point on

the ellipse to the two foci is constant, it follows that

(Sum of distances from (0, b) to foci) = (sum of distances from (a, 0) to

foci)

Example 3 – Finding the Standard Equation of an Ell

Find the standard form of the equation of the ellipse shown in Figure 4.23.

Figure 4.23

Example

From Figure 4.23, the foci occur at (–2, 0) and (2, 0).

So, the center of the ellipse is (0, 0), the major axis is horizontal, and the

ellipse has an equation of the form

Standard form, horizontal major axis

Also from Figure 4.23, the length of the major axis is 2a = 6.

This implies that a = 3.

Solution

Solution

Example

Sketch the ellipse 𝑥2 + 4𝑦2 = 16.

Solution

By dividing by 16 on both sides, the given equation 𝑥2 + 4𝑦2 = 16 is equivalent to

𝑥2

16+4𝑦2

16=16

16𝑥2

16+

𝑦2

4= 1

Thus, 𝑎 = 4, 𝑏 = 2 and𝑐 = 2 3

Because 𝑐2 = 𝑎2 − 𝑏2 = 42 − 22 = 16 − 4 = 12.

The foci are (2 3, 0) and (−2 3, 0) because the major axis is x-axis.

Solution

x-intercepts: (4,0) and (−4,0)

y-intercepts: (0,2) and 0,−2

Since the major axis is x-axis, the vertices are x-intercepts.

Example

Sketch the ellipse 4𝑥2 + 𝑦2 = 36.

Solution

By dividing by 36 on both sides, the equation 4𝑥2 + 𝑦2 = 36 is

equivalent to 4𝑥2

36+

𝑦2

36=

36

36. Thus,

𝑥2

9+𝑦2

36= 1

We know that a=6 and b=3.

𝑐2 = 𝑎2 − 𝑏2 = 62 − 32 = 36 − 9 = 27. Thus, 𝑐 = 3 3.

Since the major axis is y-axis, the foci are (0, 3 3) and 0,−3 3 .

Solution

x-intercepts: (3,0) and (-3,0)

y-intercepts:(0,6) and (0,-6).

Since the major axis is the y-axis,

the vertices are the y-intercepts.

Hyperbolas

Hyperbolas

The definition of a hyperbola is similar to that of an ellipse.

For an ellipse, the sum of the distances between the foci and a

point on the ellipse is constant, whereas for a hyperbola, the

absolute value of the difference of the distances between the foci

and a point on the hyperbola is constant.

Hyperbolas

Figure 4.25(a)

|d2 – d1| is a constant.

Hyperbolas

The graph of a hyperbola has two disconnected parts (branches).

The line through the two foci intersects the hyperbola at two points

(vertices).

The line segment connecting the

vertices is the transverse axis,

and the midpoint of the transverse

axis is the center of the hyperbola.

See Figure 4.25(b).

Figure 4.25(b)

Hyperbolas

(a) Transverse axis is horizontal (b) Transverse axis is vertical

Hyperbolas

Be careful when finding the foci of ellipses and hyperbolas. Notice that the

relationships among a, b and c differ slightly.

Finding the foci of an ellipse:

c2 = a2 – b2

Finding the foci of a hyperbola:

c2 = a2 + b2

Find the standard form of the equation of the hyperbola with foci at (–3, 0)

and (3, 0) and vertices at (–2, 0) and (2, 0), as shown in Figure 4.26.

Figure 4.26

Example

From the graph, you can determine that c = 3, because the

foci are three units from the center.

Moreover, a = 2 because the vertices are two units from the

center. So, it follows that

b2 = c2 – a2

= 32 – 22

= 9 – 4

= 5.

Solution

Because the transverse axis is horizontal, the standard

form of the equation is

Finally, substitute a2 = 22 and to obtain

Write in standard form.

Simplify.

cont’d

Standard form, horizontal transverse axis

Solution

Example

Sketch the hyperbola 4𝑥2 − 𝑦2 = 16. Find the vertices, foci and asymptotes.

Solution

𝑥 −intercepts:

4𝑥2 − 02 = 164𝑥2 = 16𝑥2 = 4

Thus, 𝑥 = ±2.

Vertices: (2,0) and (-2,0)

Solution

Asymptotes: To convert the given form to 𝑥2

𝑎2−

𝑦2

𝑏2= 1,

4𝑥2 − 𝑦2 = 16

4𝑥2

16−𝑦2

16=16

16

𝑥2

4−𝑦2

16= 1

Thus, 𝑎 = 2 and 𝑏 = 4.

Therefore, 𝑦 =4

2𝑥 and 𝑦 = −

4

2𝑥. That is, 𝑦 = 2𝑥 and 𝑦 = −2𝑥

Solution

Foci: 𝑐2 = 𝑎2 + 𝑏2 = 4 + 16 = 20𝑐 = ± 20 = ±2 5

Example

Find the standard form of the equation of the hyperbola that has vertices at (0, −3) and 0,3 , asymptotes 𝑦 = −2𝑥 and 𝑦 = 2𝑥.

Solution

Since vertices are (0,-3) and (0,3), we know 𝑎 = 3 in the standard form of

the equation 𝑦2

𝑎2−

𝑥2

𝑏2= 1.

Thus, we need to find b in 𝑦2

32−𝑥2

𝑏2= 1.

Since one of the asymptotes is 𝑦 =

2𝑥 and 𝑎 = 3, we have 𝑏 =3

2from

𝑦 =𝑎

𝑏𝑥 =

3

𝑏𝑥 = 2𝑥.

Hyperbolas

An important aid in sketching the graph of a hyperbola is the

determination of its asymptotes, as shown in below.

Figure 4.33

(a) Transverse axis is horizontal;

conjugate axis is vertical.

(b) Transverse axis is vertical;

conjugate axis is horizontal.

Hyperbolas

Each hyperbola has two asymptotes that intersect at the center of the

hyperbola. Furthermore, the asymptotes pass through the corners of a

rectangle of dimensions 2a by 2b.

The line segment of length 2b joining (0, b) and (0, –b) [or (–b, 0) and

(b, 0)] is the conjugate axis of the hyperbola.

Parabolas

ParabolasYou have learned that the graph of the quadratic function

f(x) = ax2 + bx + c

is a parabola that opens upward or downward. The following definition

of a parabola is more general in the sense that it is independent of the

orientation of the parabola.

Note in the figure above that a parabola is symmetric with respect to its

axis. Using the definition of a parabola, you can derive the following

standard form of the equation of a parabola whose directrix is parallel

to the x-axis or to the y- axis.

Parabolas

Parabolas

Notice that a parabola can have a vertical or a horizontal

axis. Examples of each are shown below.

(a) Parabola with vertical axis (b) Parabola with horizontal axis

Example

Find the focus and directrix of the parabola 𝑦 = −2𝑥2. Then, sketch the parabola.

Solution

𝑥2 = −1

2𝑦 = 4𝑝𝑦 = 4 ∙ −

1

8𝑦

Thus, 𝑝 = −1

8. The focus is

0,−1

8and directrix is 𝑦 =

1

8.

Example 2 – Finding the Standard Equation of a Parabola

Find the standard form of the equation of the parabola with vertex at the

origin and focus at (2, 0).

Figure 4.19

Example

Example 2 – Finding the Standard Equation of a Parabola

Figure 4.19

Solution

The axis of the parabola is horizontal, passing through (0, 0) and (2, 0), as shown in Figure 4.19.

So, the standard form is

y2 = 4px.

Because the focus is p = 2 units from the vertex, the equation is

y2 = 4(2)x

y2 = 8x.

Parabolas occur in a wide variety of applications. For instance, a

parabolic reflector can be formed by revolving a parabola about its axis.

The resulting surface has the property that all incoming rays parallel to

the axis are reflected through the focus of the parabola.

This is the principle behind the construction of the parabolic mirrors

used in reflecting telescopes.

Real Life Application of Parabola

Parabolas

Conversely, the light rays emanating from the focus of a parabolic reflector

used in a flashlight are all parallel to one another, as shown in Figure 4.20.

Figure 4.20