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.
Circle
Ellipse
Parabola
Basic Conics
Figure 4.16
Hyperbola
Introduction
When the plane does pass through the vertex, the resulting figure is a
degenerate conic, as shown in Figure 4.17.
Point
Line
Degenerate Conics
Figure 4.17
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
Equation of a circle
(x – h)2 + (y – k)2 = r 2.
We know that the center of a circle is at (h, k) and that
the radius of the circle is r.
Circle
standard form of the equation of a circle
Circle with center at origin
x 2 + y 2 = 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 − 𝑥2
2
+ 𝑦1 − 𝑦2
2
Thus,
𝑑=
3 − −1
2
+ 4−2
2. (x + 1)2 + (y – 2)2 = 20.
2
=
3+1
2
+ 22 =
42 + 4 = 16 + 4 = 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.
Figure 4.22
The standard form of the equation of an ellipse takes one of two forms,
depending on whether the major axis is horizontal or vertical.
Ellipses
(a) Major axis is horizontal;
minor axis is vertical.
(b) Major axis is vertical;
minor axis is horizontal.
Ellipses
In 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
Example
Find the standard form of the equation of the ellipse shown in Figure 4.23.
Figure 4.23
Solution
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
Example
Sketch the ellipse 𝑥 2 + 4𝑦 2 = 16.
Solution
By dividing by 16 on both sides, the given equation 𝑥 2 + 4𝑦 2 = 16 is
equivalent to
𝑥 2 4𝑦 2 16
+
=
16 16
16
𝑥2
16
+
𝑦2
4
=1
Thus, 𝑎 = 4, 𝑏 = 2 and𝑐 = 2 3
Because 𝑐 2 = 𝑎2 − 𝑏 2 = 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 4𝑥 2 + 𝑦 2 = 36.
Solution
By dividing by 36 on both sides, the equation 4𝑥 2 + 𝑦 2 = 36 is
4𝑥 2
𝑦2
36
equivalent to
+ = . Thus,
36
36
36
𝑥2 𝑦2
+
=1
9 36
We know that a=6 and b=3.
𝑐 2 = 𝑎2 − 𝑏 2 = 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
|d2 – d1| is a constant.
Figure 4.25(a)
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
Example
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
Solution
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
cont’d
Because the transverse axis is horizontal, the standard
form of the equation is
Standard form, horizontal transverse axis
Finally, substitute a2 = 22 and
to obtain
Write in standard form.
Simplify.
Example
Sketch the hyperbola 4𝑥 2 − 𝑦 2 = 16. Find the vertices, foci and
asymptotes.
Solution
𝑥 −intercepts:
4𝑥 2 − 02 = 16
4𝑥 2 = 16
𝑥2 = 4
Thus, 𝑥 = ±2.
Vertices: (2,0) and (-2,0)
Solution
𝑥2
𝑎22
𝑦2
𝑏2
Asymptotes: To convert the given form to − = 1,
4𝑥 2 − 𝑦 = 16
4𝑥 2 𝑦 2 16
−
=
16 16 16
𝑥2 𝑦2
−
=1
4 16
Thus, 𝑎 = 2 and 𝑏 = 4.
4
4
Therefore, 𝑦 = 𝑥 and 𝑦 = − 𝑥. That is, 𝑦 = 2𝑥 and 𝑦 = −2𝑥
2
2
Solution
Foci: 𝑐 2 = 𝑎2 + 𝑏 2 = 4 + 16 = 20
𝑐 = ± 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
𝑦2
𝑥2
the equation 2 − 2 = 1.
𝑎
𝑏
Thus, we need to find b in
𝑦2 𝑥2
− 2 = 1.
2
3
𝑏
Since one of the asymptotes is 𝑦 =
3
2𝑥 and 𝑎 = 3, we have 𝑏 = from
𝑦=
𝑎
𝑥
𝑏
=
3
𝑥
𝑏
2
= 2𝑥.
Hyperbolas
An important aid in sketching the graph of a hyperbola is the
determination of its asymptotes, as shown in below.
(b) Transverse axis is vertical;
conjugate axis is horizontal.
(a) Transverse axis is horizontal;
conjugate axis is vertical.
Figure 4.33
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)
(b, 0)] is the conjugate axis of the hyperbola.
[or (–b, 0) and
Parabolas
You have learned that the graph of the quadratic function
Parabolas
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.
Parabolas
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
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 = −2𝑥 2 . Then, sketch the
parabola.
Solution
𝑥2
1
1
= − 𝑦 = 4𝑝𝑦 = 4 ∙ − 𝑦
2
8
Thus, 𝑝 =
1
0, −
8
1
− .
8
The focus is
and directrix is 𝑦 =
1
.
8
Example 2 – Finding the Standard Equation of a Parabola
Example
Find the standard form of the equation of the parabola with vertex at the
origin and focus at (2, 0).
Figure 4.19
Example 2 – Finding the Standard Equation of a Parabola
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.
Figure 4.19
Real Life Application of Parabola
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.
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