# Conics

Document Sample

Introduction to Conics
Conic Section: Circle
Conic Section: Ellipse
Conic Section: Parabola
 What are Conic Sections?
 2 videos
1) Introduction to Conics (8 min.)Videos\Intro to Conics+Circles\Introduction to Conic Sections A.rv
2) Introduction to Circles (8:15 min.)                      Videos\Intro to Conics+Circles\Conic Sections Intro to Circles B.rv

 Circle is All points equidistant, r, from a single point, the
center.                                                   (x,y)
r
• Standard “center radius” form of a Circle?
(0,0)
Center at (0,0)

(x  k)2  ( y  k)2  r 2
Center at (h,k)
(x,y)
r
( x  h) 2 ( y  k ) 2                                                                                     (h,k)
2
      2
1                          Examples follow
r          r
1) Center (0,0) and radius 3  (x-0)2+(y-0)2=9       x2+y2=9

32

2)Center (0,2) and radius 11  (x-0)2+(y-2)2 = 121

3)Center (3,4) and radius 2  (x-3)2+(y-4)2 = 4

4)Center (-4,0) and radius 5  (x+4)2+(y-0)2 = 25

5) Center (0,0) and radius 1/2  (x-0)2+(y-0)2 = 1/4
Circles

c) Center (0,0) and radius 11   h) Center ( 2, 3) and radius 6

d)Center (0,0) and radius 5     i) Center (–3, –5) and radius 5

e)Center (2,0) and radius 6     j) Center (–11, –12) and radius 4
 What are Conic Sections?
 Video -- Introduction to Ellipse (13 min.)
Videos\Intro to Ellipses C1.rv

 Definition: All points in a plane, the sum of whose
distances from two fixed points (foci) is constant.
 The standard eq. form of an Ellipse
( x  h) 2at (h, k)k ) 2
Center  ( y            1
2            2                         ( h, k+b)
a            b
( h–a, k)               ( h+a, k)
(h,k)

( h, k–b)
( 0, +3)
2           2
x    y
   1                 ( –4, 0)                            ( 4, 0)
16   9                                     (0,0)
a
( 0, –3)
( 0, +5)
b                              x2 y2
   1            ( x  3) 2 ( y  2) 2
16 25                                    1
25         16
( –4, 0)                ( 4, 0)
c
(0,0)
( 3, 6)

( –2, 0)                                ( 8, 0)
( 0, –5)                                          (3,2)

( 3, –2)
( h, k+b)
x2 y2
a        1                  ( h–a, k)
36 25                                                   ( h+a, k)
(h,k)

x2                                       ( h, k–b)
b       y2  1
9                                                    ( h, k+a)

c ( x  1)         ( y  2) 2
2
            1
16                25                                   (h,k)

( h–b, k)                  ( h+b, k)
d
( x  2) 2 ( y  3) 2
           1
9          4
( h, k–a)
 Page 364, #s 35, 36, 39, 40, 41. 42

x2 y2
1          1
25 16

x2   y2                      4   ( x  4) 2 ( y  5) 2
2         1                                            1
144 169                               28         64

( x  4) 2 y 2            5
3                   1             4 x  9 y 2  36
9       5
 An Ellipse has 2 foci
 Definition (reworded): an Ellipse is the set of points
where the sum of the points’ distances from the 2 foci
is a constant.
 Determining the location of the 2 foci…
..\7th 5 weeks\Foci of an Ellipse C2.rv

 Important relationships:
   Let the focus length be equal to c
   c2=a2-b2                             d1                  d2
   d1+d2=2a
   Eccentricity (flatness), e = c/a,         a                 c

Examples follow
 What is the ellipse’s equation (in standard form)
given…

(-5,7)   (-3,7)       (3,7) (5,7)
 Vertices: (±5,7) Foci: (±3,7)
 c2=a2-b2
a      c
 Since, a=5 & c=3, then b=4

( x) 2 ( y ) 2        ( x) 2 ( y ) 2
2
 2 1                        1
a      b               25    16

Note: The Foci are always on the major axis !!
Sketching the
ellipse first,
might HELP !

 Vertices: (±13,1)       Foci: (±12,1)

 Vertices: (±4,7)       Foci: (±3,7)

 Vertices : (2,1), (+14,1)    Foci: (4,1), (+12,1)
b

 Vertices: (7,±5)     Foci: (7,±3)              a        c
 Page 364, #s 47 through 50 and 51 for extra credit
√         √

√          √         √

How to quickly Identify the conic from the
equation (future) ?
(x,y)
 Circle: ( x  h)  2
( y  k)    2
r
2
       2
1
r          r                             (h,k)

( x  h) 2 ( y  k ) 2                                  ( h, k+b)
 Ellipse:                         1
a 2
b 2
( h–a, k)       d1             d2       ( h+a, k)
  c2=a2-b2
 d1+d2=2a                                                    a            c
( h, k–b)
 Eccentricity (flatness), e = c/a

 Parabola:      y  a ( x  h) 2  k                                   We have studied
 if vertex is at (0,0)               y  a(x) 2                      parabolas that point
up or down (so far).
 if vertex is at ( h, k)               ( y  k )  a ( x  h) 2
Circle – set of all points that are the same
distance (equidistant), r, from a single point, the
center.

Ellipse – set of all points in a plane, the sum of whose
distances from two fixed points (foci) is constant.

Parabola – set of points in a plane that are
equidistant from a fixed line (the directrix) and a
point (the focus).
 Parabola – opening up or down, the equation is:
Point 1: And if the vertex (h,k) is at (0,0), then

y  a( x  h) 2  k becomes   x 2  4 py

Similarly if we have a parabola opening left or right
then the x and y is switched around          y 2  4 px

Point 2: p is the distance from the vertex
 to the focus and
 to the directrix
Note By the definition of a parabola the vertex is always
midway between the focus and the directrix.

 Point 3: Hence, to find that distance divide the coefficient of the
variable (the variable having a 1 as its exponent) by 4.
 Reference Drawn examples on board

SUMMARIZING…

Remember the vertex is at ( 0, 0 )…
 if the parabola opens ‘up’ then the focus is at ( 0, p)
 if the parabola opens ‘down’ then the focus is at ( 0, -p)
 if the parabola opens to the ‘right’ then the focus is at ( p, 0)
 if the parabola opens to the ‘left’ then the focus is at ( -p, 0)
x 2  ( A) y  (4 p) y

x 2  16 y    set 4p=16   and solve for ‘p’
Opens up         solved… p=4
 therefore the focus is at ( 0, 4)

1          set 4p= –1/2 and solve for ‘p’
x  y
2

2            solved… p= –1/8
Opens down        therefore the focus is at ( 0, –1/8)

y 2  9x     set 4p=9 and solve for ‘p’
 solved… p = 9/4 = 2 ¼
Opens right
 therefore the focus is at ( 2 ¼, 0)
 Page # 363, problem #s 1, 2, 3, 4, 11,12
Due Wednesday

 Page #363, problem #s 13, 14, 15, 16
Due TBD

 Page #363, problem #s 17, 18, 19, 20, 21, 22, 23
Due TBD
x 2  16 y    set 4p=16   and solve for ‘p’
Opens up         the focus is at ( 0, 4) -- see previous slide
 and the directrix, y = –4

1          set 4p= –1/2 and solve for ‘p’
x  y
2

2            the focus is at ( 0, –1/8)
Opens down        and the directrix, y = +1/8

y 2  9x     set 4p=9 and solve for ‘p’
 the focus is at ( 2 ¼, 0)
Opens right
 and the directrix, x = –2 ¼
 Eccentricity = e = c/a

 Explain what the effect is on the ellipse’s
shape as the focus’s distance from the center
(‘c’) approaches the vertex’s distance from
the center (‘a’) -- in other words, when ‘e’
approaches a value of 1.
 Please note that the next school-wide writing prompt will
take place on Tuesday, 4/5/11 during 2nd period.

The prompt is as follows:
 "The use of Cornell Notes, Flash Cards and Concept
notes and make your test preparation easier. What
other learning activities would you like to see