Slide 1 Slide 3
Equations and Graphs of Parabolas
Equ a ti on Ve rte x Foc u s Di re c trix Ax is o f s ym m e try
P ARAB OLA
F (0, c)
F (c, 0)
Slide 2 Slide 4
Equations and Graphs of Parabolas
Parabola – set of points in a plane that are equidistant
Equ a ti on Ve rte x Foc u s D ire ctri x Ax is o f S y m m e try
from a fixed point and a fixed line(not passing
through the fixed point.
Axis of
Focus - the fixed point of a parabola. Sym m et r y
F ocu s
Directrix - the fixed line of a parabola.
Axis of Symmetry – The line that goes
through the focus and is perpendicular
to the directrix.
Vertex – the point of intersection of Ver t ex Dir ect r ix
the axis of symmetry and the
parabola.
Slide 5 Slide 7
Equations and Graphs of Parabolas
Equ a ti on Ve rte x Fo c u s D ire c tri x Ax is o f s ym m e try P OINTS TO P ONDER
If t h e pa r abola open s u pwar d or down wa r d, th e
x-pa r t is bein g squ ar ed.
If t h e pa r abola open s to th e left or th e r igh t , th e
y-pa r t is bein g squ ar ed.
Th e dista n ce of th e focu s (fixed poin t) a n d th e
dir ectr ix (fixed lin e) fr om th e ver t ex is equ a l.
F (h , k+c)
Slide 6 Slide 8
Find the vertex, focus, and directrix of the given
Equations and Graphs of Parabolas equation:
Equ a ti on Ve rte x Foc u s Di re c trix Axi s of s y m m e try
x 2 4cy.
4c 16
c4
Since t he pa r a bola opens upwa r d, t he
focus is a bove t he ver t ex while t he
dir ect r ix is below t he ver t ex.
Slide 9 Slide 11
Find the equation given the vertex (3, 1) and the focus (3, 5).
Th e fir st t hin g t ha t you a r e going t o do is t o fin d c wit h t he use of dist a n ce
for m u la :
c ( x2 x1 ) 2 ( y2 y1 ) 2
Sin ce t he sign of 18 is n egat ive an d
4c 18
4c 12
t he y-pa r t is squ ar ed t h e gr a ph
opens t o t he left . Ther efor e, t h e
(3 3) 2 (5 1) 2
9 Ta ke not e t h a t t he x-coor dina t es of
c focus is at t h e left of t he ver t ex 0 ( 4) 2 t he focu s a nd th e ver t ex a r e t he sam e
c3 2
9
wh ile t he dir ect r ix is at t he r ight .
16
an d t h e coor dina t es of th e focus is
above t h e ver t ex, t hen ou r pa r abola
Focus ( ,0) opens u pwa r d.
Focus (0,3) 2
9
c 4
Subst it ut e t he va lue of c=4 a nd t he
directrix x ver t ex (3, 1) t o t he equa t ion ( x h) 2 4c( y k )
directrix y 3 2
axis of symmetry y 0
axis of symmetry x 0
Not e: If t h e x-pa r t is squ ar ed, u se
t he y-coor din at e of t h e ver t ex t hen
add/su bt r a ct c. If t he y-par t is
squ ar ed, use t he x-coor dina t e of t h e
ver t ex t h en add/subt r act c.
Slide 10
Slide 12
Slide 13 Slide 15
Determine the focus, directrix and axis of symmetry
Det er m in e t h e ver t ex, focu s, dir ect r ix, of the parabola with the given equation.
a n d a xis of sym m et r y of t h e pa r a bola
1. x 2 4 y
given t h e equ a t ion y 2 5 x 12 y 16
2. 3 y 2 24 x
y 2 12 y 5 x 16
4c 5 2
y 2 12 y 36 5 x 16 36 5 9
5
c 1.25 3. y 5( x )
4 2 2
y 2 12 y 36 5 x 20
( y 6) 2 5( x 4)
4. x 6 x 8 y 7
2
5. y 2 12 x 8 y 40
Slide 14 Slide 16
Det er m in e t h e ver t ex, focu s, dir ect r ix,
a n d a xis of sym m et r y of t h e pa r a bola
given t h e equ a t ion 5 x 2 30 x 24 y 51
24
5 x 2 30 x 24 y 51 4c
5
5( x 2 6 x 9) 24 y 51 5(9) 6
c
5( x 3) 2 24 y 96 5
5( x 3) 2 24( y 4)
24
( x 3) 2 ( y 4)
5
Slide 17 Slide 19
Find the standard equation of the parabola which
satisfies the given condition.
a.vertex (1,9), focus (3,9)
A satellite dish is in the shape of a
b.vertex (8,3) directrix x 10.5
parabolic surface. The dish is 12
c.vertex (4,2), focus ( 4,1) ft in diameter and 2 ft deep. How
d . focus (7,11), directrix x 1 far from the base should the
e. focus (7,11), directrix y 4 receiver be placed?
Slide 18 Slide 20
(a )ver t ex (1,−9), focu s (−3,−9)
An swer : (y + 9)2 = −16(x − 1) 12 (-6, 2) (6, 2)
(b) ver t ex (−8, 3), dir ect r ix x = −10.5 2
An swer : (y − 3)2 = 10(x + 8)
(c) ver t ex (−4, 2), focu s (−4,−1)
An swer : (x + 4)2 = −12(y − 2) Since the parabola is vertical and has its
(d) focu s (7, 11), dir ect r ix x = 1 vertex at (0, 0) its equation must be of the
An swer : (y − 11)2 = 12(x − 4) form: x2 = 4cy
(e) focu s (7, 11), dir ect r ix y = 4 At (6, 2), 36 = 4c(2) The receiver
An swer : (x − 7)2 = 14(y − 7.5) should be
so c = 4.5
placed 4.5 feet
thus the focus is at above the base
the point (0, 4.5) of the dish.
Slide 21
The towers of a suspension bridge
are 800 ft apart and rise 160 ft above
the road. The cable between them
has the shape of a parabola, and the
cable just touches the road midway
between the towers. What is the
height of the cable 100 ft from a
tower?
Slide 22
(400, 160)
Since the parabola is vertical
(300, h)
and has its vertex at (0, 0) its
equation must be of the form:
100
x2 = 4cy
90,000 = 1000(h )
h = 90 ft
160,000 = 4c (160) The cable would be
1000 = 4c 90 ft long at a point
c = 250 100 ft from a tower.
thus the equation is
x2 = 1000y
