10.1
Conic Sections and Parabolas
Quick
Review
1. Find the distance between (  1, 2) and (3,  4).
2. Solve for y in terms of x. 2 y  6 x
2
3. Complete the square to rewrite the equation in vertex form.
y  x  2x  5
2
4. Find the vertex and axis of the graph of f ( x)  2( x  1)  3.
2
Describe how the graph of f can be obtained from the graph
of g ( x)  x .
2
5. Write an equation for the quadratic function whose graph
contains the vertex (2,  3) and the point (0,3).
Quick
Review
Solutions
1. Find the distance between (  1, 2) and (3,  4). 52
2. Solve for y in terms of x. 2 y  6 x y   3 x
3. Complete the square to rewrite the equation in vertex form.
2
y   x  2 x  5 y  ( x  1)  4
4. Find the vertex and axis of the graph of f ( x)  2( x  1)  3.
2
2
2
Describe how the graph of f can be obtained from the graph
of g ( x)  x . vertex:(  1,3); axis:x  1; translation left 1 unit,
2
vertical stretch by a factor of 2, translation up 3 units.
5. Write an equation for the quadratic function whose graph
contains the vertex (2,  3) and the point (0,3). y 
3
 x  2  3
2
2
What you’ll learn about
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Conic Sections
Geometry of a Parabola
Translations of Parabolas
Reflective Property of a Parabola
… and why
Conic sections are the paths of nature: Any
free-moving
object in a gravitational field follows the
path of a conic
section.
Parabola
A parabola is the
set of all points
in a plane
equidistant from
a particular line
(the directrix)
and a particular
point (the focus)
in the plane.
A Right Circular Cone (of
two nappes)
Conic Sections and
Degenerate Conic
Sections
Conic Sections and
Degenerate Conic
Sections (cont’d)
Second-Degree
(Quadratic) Equations in
Two Variables
Ax  Bxy  Cy  Dx  Ey  F  0, where A, B, and C , are not all zero.
2
2
Structure of a Parabola
Graphs of x2=4py
Parabolas with Vertex
(0,0)
• Standard equation x2 = 4py
• Opens
Upward or
Downword
• Focus
(0,p)
• Directrix
y = -p
• Axis
y-axis
• Focal length
p
• Focal width
|4p|
y2 = 4px
To the Right
or to the left
(p,0)
x = -p
x-axis
p
|4p|
Graphs of y2 = 4px
Example Finding an
Equation of a Parabola
Find an equation in standard form for the parabola whose directrix
is the line x  3 and whose focus is the point (  3,0).
Example Finding an
Equation of a Parabola
Find an equation in standard form for the parabola whose directrix
is the line x  3 and whose focus is the point (  3,0).
Because the directrix is x  3 and the focus is (  3,0), the focal
length is  3 and the parabola opens to the left. The equation of
the parabola in standard from is:
y  4 px
2
y  12 x
2
Parabolas with Vertex (h,k)
• Standard equation
(x-h)2 = 4p(y-k)
(y-k)2 = 4p(x-h)
• Opens
Upward or downward
To the right or to the left
• Focus
(h,k+p)
(h+p,k)
• Directrix
y = k-p
x = h-p
• Axis
x=h
y=k
• Focal length
p
p
• Focal width
|4p|
|4p|
Example Finding an
Equation of a Parabola
Find the standard form of the equation for the parabola with
vertex at (1,2) and focus at (1,  2).
Example Finding an
Equation of a Parabola
Find the standard form of the equation for the parabola with
vertex at (1,2) and focus at (1,  2).
The parabola is opening downward so the equation has the form
( x  h)  4 p ( y  k ).
2
(h, k )  (1, 2) and the distance between the vertex and the focus is
p  4. Thus, the equation is ( x  1)  16( y  2).
2