# 9.1 Conic Sections Conic sections Circle Ellipse

```9.1 Conic Sections
Conic sections – curves that result from the intersection of
a right circular cone and a plane.
Circle
Ellipse
Parabola
Hyperbola
function of the form
y = ax2 + bx + c
a, b, c are real numbers & a 0
A quadratic Equation: y = x2 + 4x + 3
1
4
3
a = _____
b = _____
c = ______
x
-2
-1
-3
0
-4
The graph is a
parabola.
Where is the vertex?
Where is the axis of symmetry?
y
-1
0
0
3
3
Formula for
Vertex:
X = -b
2a
Plug x in to
Find y
9.2 Parabola
A 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.
Vertex Form of a Parabola
Parabola
Directrix
y  a ( x  h) 2  k
(opens up or down)
x  a(y  k )2  h
(opens right or left)
Focus
Axis of
Symmetry
Vertex
Vertex = (h, k)
Focus & Directrix = +/- 1 from vertex
4a
Focus
Parabola
Vertex
x
Plotting Points of the parabola can be done by:
• finding x/y intercepts and/or
• choosing values symmetric to the vertex
Directrix
Parabola Graphing
Step 1: Place equation in one of two vertex forms:
y  a ( x  h)  k
x  a(y  k )2  h
2
(opens up or down)
(opens right or left)
Step 2: Find the vertex at : (h, k) & determine which direction the parabola opens
Step 3: Use an x/y chart to plot points symmetric to the vertex & draw the parabola
Step 4: Find the focus/directrix displacement with 1/(4a)
Try the following Examples:
#1) y2 = 4x
#2) x2 = -16y
#3) (y + 4)2 = 12(x + 2)
#4) x2 – 2x – 4y + 9 = 0
1.1 Circles
Standard circle equation (x – h)2 + (y – k)2 = r2
center = (h, k)
Graph the circle: (x – 2)2 + (y + 3)2 = 9
Step 1: Find the center & radius
(x – 2)2 + (y – (-3))2 = 32
h is 2.
Step 2: Graph the center point
Step 3: Use the radius to find 4 points
directly north, south, east, west
of center.
k is –3.
Center = (2, -3)
r is 3.
General Form of a Circle Equation
(Changing from General to Standard Form)
Standard circle equation : (x – h)2 + (y – k)2 = r2 center = (h, k) radius = r
General Form of a circle : x2 + y2 + Dx + Ey + F = 0
Change from General Form to Standard Form
General Form:
x2 + y2 –4x +6y +4 = 0
Move constant
:
X’s & Y’s together :
x2 + y2 –4x
+6y
(x2 – 4x
Setup to Complete Square
½ (-4) = -2
½ (6) = 3
= -4
) + (y2 + 6y
) = -4 (-2)2 = 4
Complete the Square: (x2 – 4x +4 ) + (y2 + 6y +9 )= -4 +4 +9
Factor
Standard Form
(x – 2) (x –2) + (y + 3) (y + 3) = 9
(x – 2)2
+
(y + 3)2
=9
(3)2 = 9
9.3 Ellipse
Ellipse - the set of all points in a plane the sum of whose distances from
two fixed points, is constant. These two fixed points are called the foci.
The midpoint of the segment connecting the foci is the center of the ellipse.
P
F1
P
F2
A circle is a special kind of ellipse. Since we know about circles, the
equation of a circle can be used to help us understand the equation
and graph of an ellipse.
Comparing Ellipses to Circles
circle: (x – h)2 + (y – k)2 = r2
Ellipse: (x – h)2 + (y – k)2 = 1
a2
center = (h, k)
center = (h, k)
(0, 5)
Example1: Graph the ellipse:
25x2 + 16y2 = 400
(-4, 0)
(0, 3)
(0, 0)
(4, 0)
(0, -3)
(0, -5)
Finding Ellipse Foci:
Example2: Calculate/Find the foci above.
b2
We do not usually call ‘a’ and ‘b’ radii
since it is not the same distance
from the center all around the ellipse,
but you may think of them as
such for the purpose of locating
vertices located at : (h + a, k)
(h – a, k)
(h, k + b)
(h, k – b)
Major and Minor axis is determined
by knowing the longest and shortest
horizontal or vertical distance
across the ellipse.
Converting to Ellipse Standard Form
Example: Graph the ellipse: 9x2 + 4y2 – 18x + 16y – 11 = 0
Move constant :
9x2 + 4y2 –18x +16y
= 11
X’s & Y’s together :
(9x2 – 18x
9(x2 – 2x
Complete the Square:
Factor
) + (4y2 + 16y
) = 11
) + 4(y2 + 4y
) = 11
9(x2 – 2x + 1 ) + 4(y2 + 4y + 4) = 11 + 9 + 16
9(x – 1) (x –1) + 4(y + 2) (y + 2) = 36
9 (x – 1)2 + 4(x + 2)2 = 36
Standard Form
9(x – 1)2
+
36
(x – 1)2
4
4(y + 2)2
=1
36
+
(y + 2)2
9
=1
Can you
graph this?
Standard Forms of Equations of
Ellipses Centered at (h,k)
Equation
Center
Major Axis
Foci
Vertices
(x - h) 2 (y - k)2
+
=1
a2
b2
a2 > b2 and b2 = a2 – c2
(h, k)
Parallel to the x-axis,
horizontal
(h – c, k)
(h + c, k)
(h – a, k)
(h + a, k)
(h, k)
Parallel to the y-axis,
vertical
(h, k – c)
(h, k + c)
(h, k – a)
(h, k + a)
(x - h) 2 (y - k)2
+
=1
b2
a2
b2 > a2 and a2 = b2 – c2
y
Major axis
Focus (h + c, k)
Focus (h – c, k)
Vertex (h + a, k)
Focus (h + c, k)
(h, k)
Major axis
(h, k)
Vertex (h – a, k)
Vertex (h + a, k)
x
Focus (h – c, k)
x
Vertex (h + a, k)
9.4 Hyperbola
A hyperbola is the set of points in a plane the difference whose distances
from two fixed points (called foci) is constant
y
Vertex
Transverse
axis
y
Vertex
x
Focus
Center
x
Focus
Traverse Axis - line segment joining the vertices.
When you graph a hyperbola you must first locate the
center and direction of the traverse access
-- parallel or horizontal
Can you find the
traverse axis, center,
vertices and foci in
the hyperbola above?
Graphing Hyperbola
Hyperbola:
(x – h)2 - (y – k)2 = 1
a2
or
(y – k)2 - (x – h)2 = 1
b2
transverse axis: Horizontal
a2
b2
transverse axis: Vertical
center = (h, k) – BE Careful!
(Foci-displacement)2 = a2 + b2
Verticies at center traverse-coordinate +a
Asymptotes: y – k = m (x – h)
center traverse-coordinate –a
m = +/- y-displacement
x-displacement
Example1 :
( x - 3) 2 ( y - 1) 2
=1
4
25
Center: (3, 1)
Traverse Axis: Horizontal
Foci-displacement
F2 = 4 + 25 = 29
F = 29 = 5.4
C
Graphing Hyperbola (Cont.)
Hyperbola:
(x – h)2 - (y – k)2 = 1
a2
b2
or
(y – k)2 - (x – h)2 = 1
a2
transverse axis: Horizontal
b2
transverse axis: Vertical
center = (h, k) – BE Careful!
(Foci-displacement)2 = a2 + b2
Verticies at center traverse-coordinate +a
Asymptotes: y – k = m (x – h)
center traverse-coordinate –a
m = +/- y-displacement
x-displacement
Example2 :
(y – 2)2 – (x + 1)2 = 1
9
16
Center: (-1, 2)
Traverse Axis: Vertical
Foci-displacement
F2 = 9 + 16 = 25
F = 25 = 5
C
Graphing Hyperbola (Completing the Square)
Hyperbola:
(x – h)2 - (y – k)2 = 1
a2
b2
transverse axis: Horizontal
or
(y – k)2 - (x – h)2 = 1
a2
b2
transverse axis: Vertical
center = (h, k) – BE Careful!
(Foci-displacement)2 = a2 + b2
Verticies at center traverse-coordinate +a
Asymptotes: y – k = m (x – h)
center traverse-coordinate –a
m = +/- y-displacement
x-displacement
Example3 : 4x2 – y2 + 32x + 6y +39 = 0
Standard Forms of Equations of
Hyperbolas Centered at (h,k)
Equation
(x - h) 2 (y - k)2
=1
a2
b2
Center
Transverse Axis
Foci
Vertices
(h, k)
Parallel to the x-axis,
horizontal
(h – c, k)
(h + c, k)
(h – a, k)
(h + a, k)
(h, k)
Parallel to the y-axis,
vertical
(h, k – c)
(h, k + c)
(h, k – a)
(h, k + a)
b2 = c2 – a2
(y - h) 2 (x - k)2
=1
a2
b2
b2 = c2 – a2
22
1
()()
22
-=
-ab
xhyk
y
Focus (h – c, k)
22
1
()()
22
-=
-ba
xhyk
Focus (h + c, k)
Focus (h + c, k)
Vertex (h + a, k)
(h, k)
(h, k)
Vertex (h – a, k)
Vertex (h + a, k)
x
Focus (h – c, k)
Vertex (h + a, k)
x
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