Notes on Conic Sections

Notes on Conic Sec,ons

Today you will learn 3 things:

1. What conics sec,ons are.

2. How conic sec,ons are formed.

3. And how to use the Nspire to solve most conic sec,on problems

What are Conic Sec,ons?

SECTIONS How are they formed? we give geometric definitions of parabolas, ellipses, and hyperbolas and n this section Sec,ons are 2D fiare gures by the or intersec,on derive their Conic standard equations.

They calledcreated conic sections, conics, because they esult from intersecting plane as shown in Figure 1. of a plane awcone ith twith wo caones.

ellipse

parabola

hyperbola

PARABOLAS

A parabola is the set of points in a plane that are equidistant from a fixed point F (called he focus) and a fixed line (called the directrix). This definition is illustrated by Figure 2.

Notice that the point halfway between the focus and the directrix lies on the parabola; it is called the vertex. The line through the focus perpendicular to the directrix is called the axis of the parabola.

In the 16th century Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. Since then, parabolic shapes have been used n designing automobile headlights, reflecting telescopes, and suspension bridges.

See Challenge Problem 2.14 for the reflection property of parabolas that makes them so useful.)

We obtain a particularly simple equation for a parabola if we place its vertex at the origin O and its directrix parallel to the x-axis as in Figure 3. If the focus is the point !0, p",

Parabola ellipse

parabola

hyperbola

A parabola is formed when a plane intersects a only one cone and passes through the base of FIGURE 1 Conics the cone.

Form of parabola axis Equation

2 ■ REVIEW

yPARABOLAS = a(x –h)2 + k

x = a(y – k)2 + h

A parabola is the set of points in a plane that are equidistant from a fixed point the focus) and a fixed line (called the directrix). This definition is illustrated by focus F Notice that the point halfway between the focus and the directrix lies on the parab called the vertex. The line through the focus perpendicular to the directrix is c axis of the parabola.

In the 16th century Galileo showed that the path of a projectile that is shot OF CONIC SECTIONS air at an angle to the ground is a parabola. Since then, parabolic shapes have b directrix vertex in designing automobile headlights, reflecting telescopes, and suspension If we write , then 2.14 the standard equationproperty of a parabola (1) becomes a!

1$!4p"Problem y ! them ax 2. so (See Challenge for the reflection of parabolas that makes FIGURE 2 It opens upward downward if p equation Figure 4, parts (a) place and (b)].

The a p # 0a and $ 0 [seefor Weifobtain particularly simple a parabola if we its vertex graph is symmetric with respect to parallel the y-axis because (1)asisinunchanged when is replaced xfocus y gin O and its directrix to the x -axis Figure 3.

If the is the poi Focus P(x, y) by !x. then the directrix has the equation y ! !p. If P!x, y" is any point on the parabola y y y is y distance from P to the focus

Vertex

(h, k)

(h, k)

Axis of Symmetry

x=h

y=k

F(0, p)

y

(0, p)

x

0

# PF # ! sx " !y ! p" and the distance from P to the( p, 0) directrix is # y " p #. (Figure ( p, 0) 3 illustrates the ca

0

y=_p

2

y=_p

p x

O

p # 0.) xThe defining property x is that these distances x 0 of a parabola 0 are equal:

(0, p)

y=_p FIGURE 3

2

x=_p

#

sx 2 " !y ! p"2 ! y " p

#

x=_p

We get an equivalent equation by squaring and simplifying: (a) ≈=4py, p>0

Directrix

Thomson Brooks-Cole copyright 2007

FIGURE 4

¥+10x=0

#

The length p xis 2 " the distance from the vertex to the y 2 ! 2py " p 2 ! y 2 " 2py " p 2 If we interchange in (1), we obtain x and focus or ythe vertex to the d2irec,rix and equals x ! 4py

1 2 2

y

#

2 x2 " !y ! p"2 ! y " p 2(d)!¥=4px, !y " p"p<0 (c) ¥=4px, p>0

(b) ≈=4py, p<0

y ! 4px 4aequation of the parabola with focus !0, p" and directrix y ! !p is An

p=

1

2 which is an equation of the parabola with focus ! p, 0"xand x ! !p. (Interchanging ! directrix 4py x and y amounts to reflecting about the diagonal line y ! x.) The parabola opens to the right if p # 0 and to the left if p $ 0 [see Figure 4, parts (c) and (d)]. In both cases the graph is symmetric with respect to the x-axis, which is the axis of the parabola.

”_ 52 , 0’ 0

x 5 x= 2

EXAMPLE 1 Find the focus and directrix of the parabola y 2 " 10x ! 0 and sketch

the graph.

SOLUTION If we write the equation as y 2 ! !10x and compare it with Equation 2, we

FIGURE 5

see that 4p ! !10, so p ! !52 . Thus the focus is ! p, 0" ! (! 52, 0) and the directrix is x ! 52 . The sketch is shown in Figure 5.

x2 !

3 2 From triangle F1 F2 P in Figure 7 awe a 2 " c 2 2% 0. 2For convenience, let b 2 ! a 2 2 Since b 2! 2a " 2c 2 & a , it follows that 2 2 by ! x 0!. Then a y ! a b2 !or,1,iforboth x 2#a x 2 sides ! a 2,are so d

(0, b)

a

(a, 0)

x c Ellipse (c, 0) A ellipse is formed when a plane !"a, ne at 0" an are called the vertices of thex 2ell (0, _b) intersects a only o (0, b) ellipse !

3 the major axis. To find the y2-in oblique angle. is called a a y ! #b .

Equation 3 is unchanged if x is (a, 0)

2 2 about2 both axes. Notic ellipsebis2 symmetric Since ! a " c Foci= c x Major Axis = a& a , it follows tha c (c, 0) 2 ellipsex becomes circle with yand ! the 0. Then #a 2 ! 1,a or x2 ! a 2,radiu so We0"summarize follow !"a, areFIGURE called1 this the discussion vertices ofasthe ell (0, _b) is called the Conics major axis. To find the y-i y !4#b . Equation unchanged if&x2ais The ellipse From triangle FFP 3 in is Figure 7 we see that 2c , s 2 2 a " c % 0. For convenience, let b ! a " c x . Then the equati y ellipsebisx !symmetric about axes. a y ! a b or, if both sidesboth are divided by a! bNoti , PARABOLAS y 2 2 a b and the ellipse becomes a circle with radiu x y (0, b) ! !1 3 (0, a) (_a, 0) parabola a b axis A parabola is the set poi a From triangle F1 Fdiscussion weofsee 2 P in Figure 7 We summarize this as b (a, 0) 2 22 follow Minor Axis = b 2 2 2 2 has fociab !#c, where ! "axline b" ,(c "ac "%c0" 0&.,For b. The ! c focus) and aaafixed Since ! athe , itconvenience, followscthat b let -interce & x c focus 0 (_c, 0) (c, 0) 1

2

2

2 2

2

2

(0, c)

(b, 0)

FIGURE 8

(0, a) (0, _c) (0, c)

≈ ¥ + =1 b@ a@ (_c, 0)

2

2

2

2 2

2

2

2 2

2

2

2

2

2

F

2 2 2 x 2#a22 ! y ! 0.bThen 1,Notice or x !sides #a. The corresp x ! a y2 ! a 2bx 22 ! or,athat if, so both arehalfwa divide the point !"a, 0" are called the vertices of the ellipse and the line segm (0, _b) 2 called the vertex.

The line 4 is called the major2 axis. To find the2 y-intercepts2wex set x !y 22 (0, b) !or y2 i 3 y ! #b. Equation 3 is unchanged if x isparabola. replaced by 2"x axis of the a b a ellipse is symmetric about both axes. Notice that if the foci coinc b (a, 0)and the ellipse becomes a circle In with the radius 16thr ! century 2 a2! b. Gali 2b 2 ! a 2 " c 2 2& a 2, it follows that b & Since x We summarize this discussion x0 c (c, 0) as follows Figure 8). air at an angle2(see toalso the groun 2 2 2

y

(_a, 0)

2

2

The ellipse x If the foci located ony ( x −ofh)an+ ellipse ( y − k ) are ! =1 b tion by interchanging x and y in a(4). (See a b

directrix y ! 0. Then x #a ! 1, or x ! a , so x ! # in designing automobile 2 2 ellipse 2 0" are called the vertices of the The ellipse 4 !"a, (0, _b) has foci !#c, 0" , where c " b , 2 2! a x y Problem (See Challenge 2.14 for FIGURE 2 is called the major axis.

To find the y -interce ! 2 !1 a$b%0 y ais2obtain b a particularly We y ! #b .

Equation 3 unchanged if x is repla The ellipse 5 (0, a) y 2 2 FIGURE 8 axes.

Notice gin Oabout its directrix paral has ellipse foci !#c,is0"symmetric , where c2 ! aand " bboth , and vertices !#a, 0"tha .

2 2 (0, c) 2 2 ¥ ≈ and the ellipse becomes a circle with radius r !

P(x, y) xlocated y on (0, _a) + =1 then the directrix has the equ If the foci of an ellipse are y − k x − h b@ a@ (b, 0) !follows !

We summarizearethis discussion as 2y-axis 2#c"(se If the foci of an ellipse locatedfrom on= the1P at !0,focus ,t distance to the + (_b, 0) (b, 0) b a 2 2 tion by interchanging x and y in (4).

(See F(0, p) tion by interchanging x0 x a y x and y inb(4). (See Figure 9.)

4 The ellipse (0, _c) x2 y2 (0, _c) 2 2 5 The ellipsex ! !2,1 O 2 2b has foci !0, p#c", where c ! a " y a b ˘b 2 2 vertex

(

(0, _a) (0, a)

5

)

(

)

y and xthe distance from P to 1 a$b%0 2 !

2 ! a c2 ! y=_p has foci !#c,p 0" a 2 " b 2proper , and v #,bwhere 0.) The defining

The ellipse

Key to Ellipse:

Both terms are posi,ve 2

( x − h) + ( y − k ) a

2

( y−k) a

2

b

2

2

2

x − h) ( + b

2

c

=1

2

=1

And length of focus is c2 = a2-­‐b2

Circle A circle is a special type of ellipse. Just like a square is a special type of rectangle, a circle is a subset of an ellipse.

And is formed when a plane intersects with a cone and is parallel to the base of the cone.

What conic section will this equation, (x+2)2 + (y+3)2 = 36, generate?

Circle Identify the center.

( x - h )2 + ( y - k )2 = r 2 ( x - 2 )2 + ( y - 3 )2 = 16 Center is at (h, k) = (2, 3) Find the radius.

r 2 = 16 2

r = 16 r =4

â&#x20AC;˘

(2, 3)

Sketch the graph.

Thomson Brooks-Cole copyright 2007

ics

x

P (x, y)

0

F¡(_c, 0)

F™(c, 0) x

Hyperbola

FIGURE 11

a a with respect to both axes.

Hyperbolas occur frequently as graphs of equations in chemistry, physics, biology, and To analyze the hyperbola further, we look economics (Boyle’s Law, Ohm’s Law, supply and demand curves). A particularly signifi(_a, 0) (a, 0) cant application of hyperbolas is found in the navigation systems developed in World Wars x2 I and II (see Exercise 051). (_c, 0) (c, 0) x !1" 2 Notice that the definition of a hyperbola is similar to that of an ellipse; the onlya change is that the sum of distances has become a difference of distances. In fact, the derivation of 2 shows thatearlier x 2 % for a 2, an so ellipse. x ! sx the equation of a hyperbola is also similar to This the one given It is % lefta. means that the hyperbola consists of two as Exercise 52 to show that when the foci are on the x-axis at !!c, 0# and the differencepart of FIGURE 121 # PF2 ! !2a, then the equation When we draw a hyperbola distances is $ PF of the hyperbola is it is useful to fir $ $ $ ≈ ¥ lines y ! !b"a#x and y ! #!b"a#x shown in - =1 2 2 hyperbola a@ b@ parabola the asymptotes; that is, they come a x yapproach # !

1 6 a2 b2

$ $

A hyperbola is formed when a plan intersects to cones and is parallel ellipse to the axis of the cones.

P is on the hyperbola when | PF¡|-| PF™ |=!2a

b

y=_ a x

y

(_a, 0) (_c, 0)

7 The hyperbola where c 2 ! a 2 " b 2. Notice that the x-intercepts are again !a and the points !a, 0# x 2 and y # !#a, 0# are the vertices of the hyperbola. But if we put x ! 0 in Equation 6 we a 2 get b y y 2 ! #b 2, which is impossible, so there is no y-intercept. The hyperbola is symmetric with respect to both axes.(0, c) has foci !!c, 0#, where c 2 ! a 2 " b 2, vert To analyze the hyperbola further, we look at Equation 6 and y ! !!b"a#x . obtain a a

b

y= a x

y=_ b x

(a, 0) 0

(c, 0)

y= b x

(0, a) x 2

y2 ! x1 " If % the1 foci of a hyperbola are on the y-axi (0, _a)a b2 obtain the following information, which is illu

x

2

0

This shows that x 2 % a 2, so $ x $ ! sx 2 % a. Therefore, we have x % a or x $ #a. This means that the hyperbola(0, _c) consists of two parts, called its hyperbola branches.

8 The FIGURE 12 When we draw a hyperbola it is useful to first draw its asymptotes, which are the dashed y2 x ≈ ¥ A-parabola is the set of points in a plane are equidistant from a fixed point F (called 2 # FIGURE 13 that lines y ! !b"a#x and y ! #!b"a#x shown in Figure 12.

Both branches of the hyperbola a b =1 a@ focus) b@ ¥ the ≈asymptotes; that approach is,definition they come arbitrarily close to by the Figure asymptotes. the and a fixed line (called the directrix).

This is illustrated 2.

- =1 has foci !0, !c#, where c 2 ! a 2 " b 2, vert a@ b@ Notice that the point halfway between the focus and the directrix lies on the parabola; it is y ! !!a"b#x.

The hyperbola 7 called the vertex. The line through the focus perpendicular to the directrix is called the x2 y2 axis of the parabola. # 2 !1 a2 b

PARABOLAS

Thomson Brooks-Cole copyright 2007

E 1

where c 2 ! a 2 " b 2. Notice that the x-interc

derive theiry standard equations. They are called conic sections,!#a, or 0#conics, are the because vertices ofthey the hyperbola. B A hyperbola is the sety of all points in a plane the difference of whose distances from two 2 2 y ! #b , which is impossible, so there is result from intersecting a cone with a plane as shown in b b Figure 1. fixed pointsy=_ F1 and x F2 (the foci) y=is xa constant. This definition is illustrated in Figure 11.

y In the 16th century Galileo showed that the path of a projectile that is shot into the 2 (0, c) foci !!c, 0#, Since where c then, ! a2 " b 2, verticesshapes !!a, 0#, have and asymptotes air at an angle to the ground ishas a parabola. parabolic been used y ! !!b"a#x . a a x y= x y=_ in designing automobile headlights, reflecting telescopes, and suspension bridges. b b (0, a) (See Challenge Problem 2.14 for the reflection property of parabolas that makes them so useful.)

If the foci of a hyperbola are on the y-axis, then by reversing the roles of x and y we x simple equation for a parabola if we place its vertex at the ori(0, _a) We obtain0 a particularly obtain the following information, which is illustrated in Figure 13. gin O and its directrix parallel to the x-axis as in Figure 3. If the focus is the point !0, p", then the directrix !hyperbola !p. If P!x, y" is any point on the parabola, then the (0, _c)has the equation yThe 8 distance from P to the focus is y2 x2

FIGURE 13 ¥ ≈ - =1 a@ b@

a2

#

b2

!1

# PF # ! sx " ! y ! p" and the distance from P to the directrix is # y " p #. (Figure 3 illustrates the case where 2

2

has foci !0, !c#, where c 2 ! a 2 " b 2, vertices !0, !a#, and asymptotes y ! !!a"b#x.

p # 0.) The defining property of a parabola is that these distances are equal:

#

sx 2 " ! y ! p"2 ! y " p

#

We get an equivalent equation by squaring and simplifying:

#

x 2 " ! y ! p"2 ! y " p

#

2

! ! y " p"2

x 2 " y 2 ! 2py " p 2 ! y 2 " 2py " p 2

Key to Hyperbola:

One term is posi,ve, the other nega,ve.

2

2

2

2

x y − 2 =1 2 a b y x − 2 =1 2 a b And length of focus is c2 = a2+b2