Conics can be formed by the intersection
of a plane with a conical surface. If the
plane passes through the Vertex of the
conical surface, the intersection is a
Degenerate Case (a point, a line, or
two intersecting lines).
Conic Sections
There are 4 types of Conics which we will
investigate:
1. Circles
2. Parabolas
3. Ellipses
4. Hyperbolas
A circle is a set of points in the xy-plane that are a
fixed distance r from a fixed point (h, k). The
fixed distance r is called the radius, and the fixed
point (h, k) is called the center of the circle.
y
(x, y)
r
(h, k)
x
The standard form of an equation of a
circle with radius r and center (h, k) is
 x  h
2
 y  k  r
2
2
Graph (x 1)  ( y  3)  16 by hand.
2
2
( x  1)  ( y  3)  16
2
2
( x  ( 1))  ( y  3)  4
2
( x  h)  ( y  k )  r
2
2
2
2
2
h = -1, k = 3, r = 4
Center: (-1, 3), Radius: 4
(-1, 7)
y
(3,3)
(-5, 3)
(-1,3)
x
(-1, -1)
The general form of the equation of
a circle is
x  y  ax  by  c  0
2
2
Find the center and radius of
2
2
x  y  4 x  8 y  5  0.
x  4x  y  8y  5
2
2
x  4x_ y  8y_  5
2
2


2
  4  4
 
 2 




2
 8  16
 
 2
x  4 x  4  y  8 y  16  5  4  16
2
2
 x  2
2
  y  4  25
2
Center: (2, -4), Radius: 5
Find the center, the radius and graph:
x2 + y2 + 10x – 4y + 20 = 0
Find the equation of the circle with
radius 3 in QI and tangent to the
y-axis at ( 0 , 2 )
Find the equation of the circle with
center at ( 2 , -1 ) through ( 5 , 3 )
Find the equation of the circle with
endpoints of the diameter at
( 3, 5 ) and ( 3 , 1 )
Find the equation of the circle that
Goes through these 3 points:
(3, 4), (-1, 2), (0, 3)
A parabola is defined as the collection of all
points P in the plane that are the same
distance from a fixed point F as they are
from a fixed line D. The point F is called the
focus of the parabola, and the line D is its
directrix. As a result, a parabola is the set of
points P for which
d(F, P) = d(P, D)
The standard form of the equation of a
parabola with directrix parallel to the
2
y-axis is
(opens left or
( y  k )  4 p( x  h)
right)
The standard form of the equation of a
parabola with directrix parallel to the
x-axis is
(opens up or
2
down) ( x  h)  4 p( y  k )
Where (h, k) represents the vertex of the parabola and “p”
represents the distance from the vertex to the focus.
The Axis of Symmetry is the line through
which the parabola is symmetrical.
The Latus Rectum is a line segment
perpendicular to the Axis of Symmetry
through the focus with endpoints on
the parabola. The length of the Latus
Rectum is “4p”.
The Latus Rectum helps define the
“width” of the parabola.
6
4
2
AXIS OF SYMMETRY
parabola
B
-5
latus rectum
-2
Directrix
focus C
VERTEX
-4
D
5
Parabola with Axis of Symmetry Parallel to xAxis, Opens to the Right, a > 0.
Equation
2
 y  k   4a  x  h
Vertex
(h, k)
Focus
(h + a, k)
Directrix
x = -a + h
y D: x = -a + h
Axis of
symmetry
y=k
V = (h, k)
F = (h + a, k)
x
Parabola with Axis of Symmetry Parallel to xAxis, Opens to the Left, a > 0.
Equation
2
 y  k   4 a  x  h 
Vertex
(h, k)
Focus
(h - a, k)
D: x = a + h
y
Axis of
symmetry
y=k
Directrix
x=a+h
F = (h - a, k)
x
V = (h, k)
Parabola with Axis of Symmetry Parallel to yAxis, Opens up, a > 0.
Equation
2
 x  h  4a  y  k 
y
Vertex
(h, k)
Focus
(h, k + a)
Directrix
y = -a + k
Axis of symmetry
x=h
F = (h, k + a)
V = (h, k)
D: y = - a + k
x
Parabola with Axis of Symmetry Parallel to yAxis, Opens down, a > 0.
Equation
2
 x  h   4 a  y  k 
y
Vertex
(h, k)
Focus
(h, k - a)
Directrix
y=a+k
Axis of symmetry
x=h
V = (h, k)
D: y = a + k
F = (h, k - a)
x
Find an equation of the parabola with vertex
at the origin and focus (-2, 0). Graph the
equation by hand and using a graphing utility.
Vertex: (0, 0); Focus: (-2, 0) = (-a, 0)
y  4ax
2
y  4(2) x
2
y  8 x
2
The line segment joining the two points
above and below the focus is called the
latus rectum.
Let x = -2 (the x-coordinate of the focus)
2
y  8 x
2
y  8( 2)
y  16
2
y  4
The points defining the latus rectum are (-2, -4)
and (-2, 4).
10
(-2, 4)
(0, 0)
10
(-2, -4)
10
0
10
Parabolas: Example Problems
Write the equation of a parabola with
vertex ( 0 , 0 ) and focus ( 2, 0 )
Parabolas: Example Problems
Find the focus, directrix, vertex,
and axis of symmetry.
y2 – 12x – 2y + 25 = 0
Parabolas: Example Problems
Write the equation of the parabola
with focus ( 0 , -2 ) and directrix x = 3
Parabolas: Example Problems
Find the focus, directrix, vertex,
and axis of symmetry.
x2 + 4x + 2y + 10 = 0
Parabolas: Example Problems
Write the equation of the parabola
with vertex ( 4 , 2 ) and directrix y = 5
Parabolas: Example Problems
Write the equation of the parabola
with directrix y = 3 and focus ( 3 , 5 )
Parabolas: Example Problems
Find the focus, directrix, vertex,
axis of symmetry, and length of the
latus rectum.
x2 – 4x – 12y – 32 = 0
Parabolas: Example Problems
Write this equation of a parabola
in standard form: y 2  3x  6 y  0
Find the vertex, focus and directrix of
2
x  4x  8y  20  0.
x  4 x  8 y  20  0
2
x  4 x  8 y  20
2
x  4 x  _  8 y  20
2





2
 4  4
 
 2
x  4 x  4  8 y  20  4
2
 x  2
2
 8 y  3
 x  2  8 y  3
2
x  h 
2
 4 p y  k 
Vertex: (h, k) = (-2, -3)
p=2
Focus: (-2, -3 + 2) = (-2, -1)
Directrix: y = -2 + -3 = -5
10
(-6, -1)
0 (2, -1)
10
y = -5
(-2, -3)
10
(-2, -1)
10
An ellipse is the collection of points in the plane the
sum of whose distances from two fixed points,
called the foci, is a constant.
Minor Axis
y
P = (x, y)
Focus1
Focus2
Major Axis
x
The standard form of the equation of an
ellipse with major axis parallel to the
x-axis is
2
2
( x  h) ( y  k )


1
2
2
a
b
The standard form of the equation of an
ellipse with major axis parallel to the
y-axis is
( x  h) ( y  k )


1
2
2
b
a
2
2
(h,k) is the center of the ellipse
For any ellipse,
a b c
2
2
2
“2a” represents the distance along the
major axis (a is always greater than b)
“2b” represents the distance along the
minor axis
“c” represents the distance from the center
to either focus (the foci of an ellipse are
always along the major axis)
Ellipse with Major Axis Parallel to the x-Axis
where a > b and b2 = a2 - c2.
Equation
2
2
 x  h   y  k   1
2
2
a
b
y
(h - c, k)
Center
(h, k)
Foci
Vertices
(h + c, k) (h + a, k)
(h + c, k)
Major axis
(h - a, k)
(h, k)
(h + a, k)
x
Ellipse with Major Axis Parallel to the x-Axis
The ellipse is like a circle, stretched more
in the “x” direction
y
Focus 2
Focus 1
Major axis
(h, k)
x
Ellipse with Major Axis Parallel to the y-Axis
where a > b and b2 = a2 - c2.
Equation
2
2
 x  h   y  k   1
2
2
b
a
y
Center
(h, k)
Foci
Vertices
(h, k + c) (h, k + a)
(h, k + a)
(h, k + c)
(h, k)
(h, k - c)
x
Major axis
(h, k - a)
Ellipse with Major Axis Parallel to the y-Axis
The ellipse is like a circle, stretched more
in the “y” direction
y
Focus 1
(h, k)
Focus 2
x
Major axis
Ellipses: Example Problems
Sketch the ellipse and find the center,
foci, and the length of the major and
minor axes:
x  42  y  52
25

16
1
Ellipses: Example Problems
Find the center and the foci.
Sketch the graph.
9 x  4 y  18x  16 y  11  0
2
2
Ellipses: Example Problems
Write the equation of the ellipse with
center ( 0 , 0 ), a horizontal major axis,
a = 6 and b = 4
Ellipses: Example Problems
Write the equation of the ellipse with
x-intercepts  2 and y-intercepts  3
Ellipses: Example Problems
Write the equation of the ellipse with
foci ( -2 , 0 ) and ( 2 , 0 ), a = 7
Ellipses: Example Problems
6
Write the equation of this ellipse
4
2
-5
5
-2
-4
-6
Ellipses: Example Problems
Find the center, foci, and graph the
ellipse:
16x2 + 4y2 – 96x + 8y + 84 = 0
Ellipses: Latus Rectum
The length of the Latus Rectum for an
Ellipse is
2
Latus
Rectum
2b
a
Latus
Rectum
(h, k)
By knowing the Latus Rectum, it makes the graph
of the ellipse more accurate
Ellipses: Latus Rectum
Use the length of the latus rectum in
Graphing the following ellipse:
( x  5) ( y  1)

1
36
16
2
2
Find an equation of the ellipse with center at the
origin, one focus at (0, 5), and a vertex at (0, -7).
Graph the equation by hand
Center: (0, 0)
Major axis is the y-axis, so equation is of the form
2
2
x
y
 2 1
2
b a
Distance from center to focus is 5, so c = 5
Distance from center to vertex is 7, so a = 7
b  a  c  7  5  49  25  24
2
2
2
2
2
2
2
x
y
2
and
a

7
,
b

24


1
2
2
b a
2
2
2
2
x
y
 2 1
24 7
x
y

1
24 49
(0, 7)
FOCI
5
(  24,0)
( 24,0)
5
0
5
(0, -7)
5
Find the center, major axis, foci, and vertices of
4 x 2  9 y 2  32 x  36 y  64  0
4 x  32 x  9 y  36 y  64
2
2
4 x 2  8 x  _   9 y 2  4 y  _   64







2
2
 8  16
  4  4
 
 
 2
 2 
4 x 2  8 x  16  9 y 2  4 y  4  64  64  36
4 x  4  9 y  2  36
2
 x  4
9
2
2
y  2


4
2
1
 x  4   y  2  1
9 2
4 2
 x  h   y  k   1
2
a
2
2
b
2
Center: (h, k) = (-4, 2)
Major axis parallel to the x-axis
c  a b 94 5
2
2
2
Vertices: (h + a, k) = (-4 + 3, 2) or (-7, 2) and (-1, 2)
Foci: (h + c, k) = ( 4  5, 2) or
( 4  5, 2) and ( 4  5, 2)
(-4, 4)
4
V(-7, 2) F(-6.2, 2) F(-1.8, 2)
2
C (-4, 2)
8
6
4
2
V(-1, 2)
0
(-4, 0)
2
4
A hyperbola is the collection of points in the
plane the difference of whose distances from
two fixed points, called the foci, is a constant.
The standard form of the equation of a
hyperbola with transverse axis parallel to
x-axis is
2
2
( x  h) ( y  k )


1
2
2
a
b
The standard form of the equation of a
hyperbola with transverse axis parallel to
y-axis is
( y  k ) ( x  h)


1
2
2
a
b
2
2
(h,k) is the center of the hyperbola
For any hyperbola,
c  a b
2
2
2
“2a” represents the distance along the
transverse axis
“2b” represents the distance along the
conjugate axis
“c” represents the distance from the center
to either focus (the foci of a hyperbola are
always along the transverse axis)
The length of the Latus Rectum for a
Hyperbola is
2
2b
a
The equations of the asymptotes for
the hyperbola are
b
these if there is a ( y  k )   ( x  h)
a
Horizontal
Transverse Axis
or these
if there is a
Vertical
Transverse Axis
a
( y  k )   ( x  h)
b
Hyperbola with Transverse Axis Parallel to the x-Axis
Latus
Rectum
Hyperbola with Transverse Axis Parallel to the y-Axis
Latus
Rectum
Find an equation of a hyperbola with center at
the origin, one focus at (0, 5) and one vertex at
(0, -3). Determine the oblique asymptotes.
Graph the equation by hand and using a
graphing utility.
Center: (0, 0)
Focus: (0, 5) = (0, c)
Vertex: (0, -3) = (0, -a)
Transverse axis is the y-axis, thus equation
is of the form
2
2
y
x
 2 1
2
a b
2
2
y
x
 2 1
2
a b
a  9, c  25
2
2
b  c  a = 25 - 9 = 16
2
2
2
2
2
y
x
 1
9 16
Asymptotes:
a
3
y x x
b
4
2
2
y
x
 1
9 16
3
y x
4
V (0, 3)
5
F(0, 5)
(-4, 0)
(4, 0)
5
0
5
V (0, -3)
F(0, -5)
5
3
y x
4
Find the center, transverse axis, vertices,
2
2
4
x

16
x

y
 8 y  16  0.
foci, and asymptotes of
4 x 2  16 x  y 2  8 y  16  0
4 x 2  4 x    y 2  8 y   16
4 x 2  4 x  _    y 2  8 y  _   16








2
2
 4  4
  8  16
 
 
 2
 2
4 x 2  4 x  4   y 2  8 y  16  16  16  16
4 x  2   y  4  16
2
2
 x  2   y  4  1
4
16
2
2
 x  2   y  4  1  x  h  y  k 

1
2
2
2
2
4
2
16
a
2
b
Center: (h, k) = (-2, 4)
Transverse axis parallel to x-axis.
a  4, b  16, c  a  b  4  16  20
2
2
2
2
2
a  2, b  4, c  2 5
Vertices: (h + a, k) = (-2 + 2, 4) or (-4, 4) and (0, 4)
Foci: (h  c, k )  ( 2  2 5, 4) or
( 2  2 5, 4) and ( 2  2 5, 4)
(h, k) = (-2, 4)
a  2, b  4, c  2 5
b
Asymptotes: y  k    x  h
a
4
y  4    x  ( 2)
2
y  4  2 x  2
y - 4 = 2(x + 2)
y - 4 = -2(x + 2)
(-2, 8)
10
V (-4, 4)
F (-6.47, 4)
10
V (0, 4)
C(-2,4)
(-2, 0) 0
F (2.47, 4)
10
Hyperbolas: Example Problems
Write the equation of the hyperbola
with center ( 4 , -2 ) a focus
( 7 , -2 ) and a vertex ( 6, -2 )
Hyperbolas: Example Problems
Find the center, foci, and
graph the hyperbola:
x2 y2

1
16 25
Hyperbolas: Example Problems
Find the center, foci, the length of
The latus rectum, and graph the
hyperbola:
2
2
x  2
( y  3)
25

9
1
Hyperbolas: Example Problems
Find the center, foci, and vertices:
16x2 – 4y2 – 96x + 8y + 76 = 0
Equilateral Hyperbolas
Equilateral Hyperbola: A hyperbola
where a = b.
When we have an equilateral hyperbola
whose asymptotes are the coordinate
axes, the equation of the hyperbola looks
like this: xy = k.
This type of hyperbola is called a
rectangular hyperbola, and is easier to
graph because the asymptotes are the x
and y axes.
Rectangular Hyperbolas
6
The equation of an rectangular hyperbola
is xy = k (where k is a constant value).
4
2
-5
5
-2
If k >0, then your graph looks like this:
-4
-6
6
4
2
If k<0, then your graph looks like this:
-5
5
-2
-4
-6
Rectangular Hyperbolas
Example: Graph by hand the hyperbola: xy = 6.
6
4
2
-5
5
-2
-4
-6
The General form of the equation of any
conic section is…
Ax  Bxy  Cy  Dx  Ey  F  0
2
2
Where A, B, and C are not all zero (however, for all of the examples
we have studied so far, B = 0).
If A = C, then the conic is a…
If either A or C is zero, then we have a…
If A and C have the same sign, but A does not equal C,
then the conic is a…
If A and C have opposite signs, then we have a ….
Conic Sections: Eccentricity
Let D denote a fixed line called the directrix; let F
denote a fixed point called the focus, which is not
on D; and let e be a fixed positive number called
the eccentricity. A conic is the set of points P in
the plane such that the ratio of the distance from
F to P to the distance from D to P equals e. Thus,
a conic is the collection of points P for which
d F ,P 
e
d D ,P 
Conic Sections: Eccentricity
To each conic section (ellipse,
parabola, hyperbola, circle) there is a
number called the eccentricity that
uniquely characterizes the shape of the
curve.
Conic Sections: Eccentricity
If e = 1, the conic is a parabola.
If e = 0, the conic is a circle.
If e < 1, the conic is an ellipse.
If e > 1, the conic is a hyperbola.
Conic Sections: Eccentricity
For both an ellipse and a hyperbola
c
e
a
where c is the distance from the center
to the focus and a is the distance from
the center to a vertex.
Conic Sections: Eccentricity
Find the eccentricity for the following
conic section: 4y2 – 8y + 9x2 – 54x + 49= 0
Conic Sections: Eccentricity
Find the eccentricity for the following
conic section: 6y2 – 24y + 6x2 – 12= 0
Conic Sections: Eccentricity
Write the equation of the hyperbola
with center ( -3 , 1 ) focus ( 2 , 1 ) and
e = 5/4
Conic Sections: Eccentricity
Write the equation of an ellipse with
center ( 0 , 3 ), major axis = 12, and
eccentricity =2/3
Conic Sections: Eccentricity
Write the equation of the ellipse and
find the eccentricity, given it has foci
( 1 , -1 ) and ( 1 , 5 ) and goes through
the point ( 4, 2 )
Conic Sections: Eccentricity
Find the center, the foci, and eccentricity.
EX 1:
4x2 + 9y2 = 36
EX 2:
4y2 – 8y - 9x2 – 54x + 49 = 0
EX 3:
25x2 + y2 – 100x + 6y + 84 = 0
Conic Sections: Solving Systems of Equations Graphically
Solve the following System of Equations
by Graphing.
9x2 + 9y2 = 36
Y – 4x = 5
Conic Sections: Solving Systems of Equations Graphically
Solve the following system of equations
by Graphing.
x2 = -4y
5x2 + y2 = 25
Conic Sections: Solving Systems of Equations Graphically
Graph the following System, then state
a sample solution.
x  16  y
2
2
4 x  9 y  36
2
2
Conic Sections: Solving Systems of Equations Graphically
Graph the following System, then state
a sample solution.
9( x  1)  4 y  36
xy  4
2
2
Theorem Identifying Conics without
Completing the Square
Excluding degenerate cases, the equation
Ax  Cy  Dx  Ey  F  0
2
2
where either A  0 or C  0:
(a) Defines a parabola if AC = 0.
(b) Defines an ellipse (or a circle) if AC > 0.
(c) Defines a hyperbola if AC < 0.
Identify the equation without
completing the square.
3x  4 y  x  9 y  10  0
A  3, C  4
2
2
AC  12  0
The equation is a hyperbola.
The standard form of an equation of a
circle of radius r with center at the
origin (0, 0) is
x y r
2
2
2