Sullivan Algebra and
Trigonometry: Section 10.2
The Parabola
Objectives of this Section
• Find the Equation of a Parabola
• Graph Parabolas
• Discuss the Equation of a Parabola
• Work With Parabolas with Vertex at (h,k)
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 equation of a parabola with vertex at
(0,0), focus at (a,0), and directrix x = -a is
y2 = 4ax
d(P,D) P
D: x = -a
d(F,P)
F = (a,0)
d(F,P) = d(P,D)
Find the equation of a parabola with vertex at
(0,0) and focus at (4,0). Graph the equation.
The distance from the vertex to the focus is
So the equation of the parabola is y2 = 16x
x = -4
(1,4)
(4,0)
a = 4.
Discuss the equation: y2 = 10x
The equation is of the form y2 = 4ax,
where 4a = 10, so a = 5/2. So, the
graph of the equation is a parabola with
vertex (0,0), a focus at the point (5/2, 0)
and directrix x = -5/2
Equations of a Parabola: Vertex at (0,0);
Focus on Axis
Vertex
Focus
Directrix Equation
y2 = 4ax
Description
(0,0)
(a,0)
x = -a
Parabola,
symmetric on x
axis, opens right
(0,0)
(-a,0)
x=a
y2 = -4ax Parabola,
symmetric on x
axis, opens left
(0,0)
(0,a)
y = -a
x2 = 4ay
(0,0)
(0,-a)
y=a
x2 = -4ay Parabola,
symmetric on y
axis, opens down
Parabola,
symmetric on y
axis, opens up
Find the equation of a parabola with focus
(0, -3) and directrix the line y = 3.
This parabola will have a vertex at (0,0),
since that point is the midpoint between the
directrix and the focus. Since the focus is
on the negative y axis with a = 3, the
equation of the parabola is
x2 = -4(3)y or x2 = -12y
Parabolas With Vertex at (h,k); Axis of
Symmetry Parallel to a Coordinate Axis
Vertex
Focus
Directrix
Equation
(h,k)
(h+a, k)
x=h-a
(y - k)2 = 4a(x - h)
(h,k)
(h-a, k)
x=h+a
(y - k)2 = -4a(x - h)
(h,k)
(h, k+a)
x=k-a
(x - h)2 = 4a(y- k)
(h,k)
(h, k - a)
x=k+a
(x - h)2 = -4a(y- k)
Find the equation of a parabola with vertex at
(-2, 3) and focus at (0, 3). Graph the equation.
The vertex and focus both lie on the horizontal line
y = 3 (the axis of symmetry). This distance from the
vertex to the focus is a = 2.
Since the focus lies to the right of the vertex, the
parabola opens to the right. The equation has the form:
(y - k)2 = 4a(x - h)
where (h,k) = (-2, 3) and a = 2
(y - 3)2 = 4(2)(x - (-2))
(y - 3)2 = 8(x + 2)
(0, 7)
D: x = -4
F = (0,3)
Axis of Symmetry: y = 3
(0, -1)
(y - 3)2 = 8(x + 2)
Find the vertex and focus of the following
parabola:
x2 + 8x = 4y - 8
First, use the method of completing the square
involving the variable x to rewrite the parabola.
x2 + 8x + ____
16 = 4y - 8 + ____
16
(x + 4)(x + 4) = 4y + 8
(x + 4)2 = 4(y + 2)
The equation is of the form (x - h)2 = 4a(y- k)
Vertex: (h,k) = (-4, -2); Focus: (h, k + a) = (-4, -1)