Si a = ( a1 , a2 ) , a
= 2, a1
a2
= 4, Hallar a
( dos soluciones)
1. PRIMERA FORMA Si a = ( a , a ) 1
2 2
( a)1 (+ ) a2 2
2
a1 = 16 a2
=2
( a) 1 2 =( 16 ) a2
( a)1 2 (+ ) a2 2 = 4
( a) 1 2 = (4)− a2 ( a1 ) 2 = 4 −
2
4 − ( a)2
2
4 = 17( a2 )
64
2
8 17
8 , 2
17
a = −
17
8 17
SEGUNDA FORMA a1
= 4a2 ⇒ a = ( 4a 2 , a) 2 ⇒ ( ) a2
a
=
a
16 + 1
( 4,1) a = a. = 17 a . 2 17 a K
W
a
=
17
4,1
,−
2
17
2
2
(a )
( a ) = 17
a =
=( 16 ) a2
4 17
1
( a1 ) = ±
2
2
=±
17
2
Donde: W
=
a
=
∴
( 4,1)
es unitario
17
k .w
17a 2
=
⇒k =±a
k
= ±2 2
( a2 ) = ±
17
⇒ a1 = 4 ±
⇒a =± 1 17 2
8 17
Por lo tanto existen dos soluciones posiles para
8 , 2
a =
17
a = −
17
P
!. De"ostrar #ue si
0
se$"ento #ue %a de
8 17
2
,−
a
17
P
≠
1
entonces los puntos #ue trisecan el
P0 P1
a
tiene la &or"a:
( P0 + 2 P1 ) / 3
'
( 2 P0 + P1 ) / 3
SO(U)I*N
→
→
P0 Q =
P1 P0
Q − P0
= P0 +
Q=
→ →
RP1
3
4 P0
P0
− P1
P1
3
− P1 3
Q=
−R=
P1
=
P0 P1 3
− P01 3
2 P1 − P0 3
PQ
+. ,allar la lon$itud de la "ediana del lado %rtices son:
( =) P = ( 3),7 , Q
− 4(,0 ), R = 1,−4
en el tri-n$ulo cu'os
.
x=
y
=
−4+3 2
0+7 2
=
M ( x, y )
=−
1 2
7 2
1 7 = − , 2 2
∴ La longitud de la mediana es : M
2
L=
1 − 1 + 7 + 4 2 2
1 7 = − , ∧ R = (1,−4) 2 2
2
MQ = PM 2
2
1 15 L = 2 + 2 = 7.52 ( − 4,)0 ( − ) x(, y ) =( ) x, y − 3,7 ( − 4 − x(),− y = ) x − 3, y − 7
−4− x = x−3 − y = y −7 − 1 = 2x 7 = 2y x=−
1
y
2
"i a
/.
=7
2
= ( m),2m , !//a , (a −) b =
2m, p a − b
= 20. Calcular b donde m ≠ 0.
∗ b =) r ( a (b1 , b2 ) (= r)
# m)(1 − r 1− r 1
( ) ( a − r a = 2m, p ( )( () ) m,2m − r m,2m = 2m, p ( m − rm,2m −( 2) mr ) = 2m, p
m,2m
= 2m
# 2 m 1 − 2r
()
=2
2 m (1 + 2 )
r
6m
− = a − b = 2a
=p =p
p
=
(m ) ,2(m − −) m,−2m = 2a ( m,2) m ( 2 − −) m,−2m
( 2) m 2( )+ 6m
2
2
= 2a ⇒
= 2a
+ 36m 2 = 20 40m = 20
4m
2
m=
10 10
⇒ b = r ( m,2m ) 10 , 20 = − 10 ,− 20 b = −1 10 10 10 10
= ( a,−) b ,( v =)
u
0. Dado los %ectores
r ( )v
# u // v
u =) r ( v
( a() ),−b = r 2b, c rem%la&amos r
2b, c( ), u + v
(
=
1$ c c
∧ 2b = c
=
+ v( =)
1,1 , "i u // v . ab / c
1,1
( ) (( )
2br , cr
2b, c
=
1,1
)()
2br + 2b, cr + c
=
1,1
+
) ( ) ( a,−b = r 2b,2b
# 2br + 2b = 1
) (
( ()a),−b = 2br ,2br
2b r + 1
)(
a = 2br
− b = 2br
a = −b
− 12 = r
=1
1 $ c + 1 = 1 c
2b
2b c
=1
#c r +1 r +1 =
1
c
r
= 1 −1
r
= 1 $c c
c
2! = c
Re"plaa"os: r
Re"plaa"os c2!
1$ c
=
2b = c
c
− 1 = 1$ c 2
! =1
c
− c = 2 − 2c c
Calculamos : ab
=2
c
=
− b #b 2
=−
1 2
6.
En el exagono regular A , B ,C , D , E , F de lado 4 unidades , hallarl a longuitud de ´s , sabiendo que :
´s =
R
´ + ED ´ + AD
1
1
3
2
2
(
)
´ EB
F A B
E D C
Solucion :
´s =
´s =
2 3 2 3
(
D− A +
(
. D− A +
D
5E
3
6
´s = −
−
1 2
)
(E−D) + E−B
E 2
− D + B− E
2A 3
)
2
2
+
2
2
B 2
´s = (− 2,−2 √ 3 )−2 ( 2,2 √ 3 ) + ( 3 B −5 E ) 3
´s =
6
(−2,−2 √ 3 )−( 4,4 √ 3 ) 3
+
3 ( 4,0 )− 5 (− 4,0 ) 6
( ) ( ) ´s = − 6,6 √3 + 32,0 3
´s =(
10 3
6
,− 2 √ 3 ) .
6.
Sean a y b vectores ℜ 2 tal que b es el opuesto a. Si b tiene el mismo sentido que el vector
= ( − 1 3,1 4) y a = 5, Hallar el vector x = 2b + a.
c
c´ =
(− ) ‖´‖= 1 1
3
,
4
y a
´ +´a . 5, hallarel vector ´x = 2 b
Solucion
θ= tan
−1
:
(− ) 3
4
θ=−37 °
a´ ´ =‖a´ ‖‖a´ ‖
∴a
a´ = 5. ( cos −37 , sin −37 ) a´ = 5
(
4 5
,−
3 5
)
a´ = ( 4,− 3 )
´ es elo! uesto de ´asera : Coo el vector b b´ = (−4,3 ) ⟹x ´=
(−4,3 )
3. En la 4$ura ad5unta6 OA7) es un cuadrado6 P6 86 R6 ' S son puntos #", "!, !C C
"edios de los lados
respecti%a"ente. )alcular
S$ + !H
. PQ y H es punto medio de '
Si 9 es punto "edio de
Hallamos : $ P$
= ( x, )y ( =) 1,3
= $Q
( x, )y ( − ) 0(,2 ) =( 2,)4 − x, y ( x, y ()− 2 = 2) − x,4 − y #x = 2− x x =1
# " = SP + P$ "
= ( 0,2) ( − )2,(0) +( 1,)3 −
"
= ( − 1,3)
Aora alla"os:
!H
!H
= !R + RH
!H
= ( 0,−)2( + ) − 1,1
!H
= ( − 1,−1)
,alla"os H
H
= ( x1 , y) 1 ( =)
RH
0,2
3,3
= HQ
( x , y) ( −) 4( ,2) (= 2,)4 − x , y 1
1
1
1
( x1 − 4, y( 1) − 2 = ) 2 − x1 ,4 − y1 # x1 − 4 = 2 − %
x1
=3
# y1 − 2 = 4 − y1
y1
=3
#y−2= 4− y =3
S$ + !H
Aora calcula"os:
( $) 1,3( + −) 1( ,−1 ) = − 2,2 ⇒ 2 2 + 2 2 = 4
;. Si
a
es un %ector unitario de
b
= (b1 , b2 )* θ =0 a1.b
b1 + b2
= 31*
= 41
) (1,0 (b1 , b2 ) = 41 ) ( b1 = 41
b1 , b2
ℜ2
6 la su"a de
= 41
a =u
= ( Cosθ , Senθ )
a = 1,0
∴ b1 + b2 = 31 41 + b2
= 31
b2 = 10
(os %ectores son: a ∧ b ⇒ a = (1),0 ∧( b
= )41,−10
<. Sea el rect-n$ulo A7)D de -rea consecuti%os son
" = ( − 2,5) &
"is"o sentido del %ector
'
48u 2
! = ( 2,1)
= ( 5,1)
' cu'os dos %rtices . Si la dia$onal
"C
tiene el
. ,allar los %rtices ) ' D.
" = b.' a.b = 48
"C // ( 5,1) "C
=
25 + 1 =
26
5 1 & = , 26 26 "! = ( 2,)1( − −) 2,5 "! = ( 4,−4 ) "!
=
16 + 16
=4
2
a=4 2
REMP(A=AMOS: a.b = 48 4 2.b = 48
b=
12 2
⇒ C = ! + b.u ⊥ 12
−
∧ 1
C)
= ( 2,1 +
C)
= ( 2,1 + − 6 13 , 13
60
C)
6 13 = ( 2,1 + − , 13
60
2
26
,
5
) (
26
= + + b.u ⊥
=
−
− 2,5 +
12
−
12 60 , 4 13 4 13
−
12 60 , 4 13 4 13
)( 52
=
− 2,5 +
12
)( 52
=
− 2,5 +
12
2
2
2
1 26
,
5
26
C)
= ( 2,1 + − 6 13 , 13
C
= ( 0.)34,-.32
)( 13
30
(
=
)
− 2,5 +
− 2
12
3 13
,
13
15
= $ 1.17,-.16
1>. Sea el rect-n$ulo A7)D6 una de cu'as dia$onales tiene por extre"os los puntos A?+6/@ ' )?<61@. Si los lados de la "a'or lon$itud son paralelos al %ector ?161@. )alcular el -rea de dico rect-n$ulo6
∧
" )// (1,1 "
=
) (
!C // 1,1 !C
2
=
2
= 1 , 1
u
"C
2
2
= ( -,16) ( −)
3,4
= ( 6,12) "C
36 144
=
( 6 )5( =) 2
2
= 2
+ ( "! ) 2
36 # 5 = 2 + ( "! ) 178 = ( "! )
"!
6 5
+
=
2
2
178
" = ( 2)(
178 )
" = 18.87u 2
11. Dado el ex-$ono re$ular A7)DEF dela 4$ura cu'o lado "ide 1> ' el %ector
M
= !C + )C + !(
M
. )alcular
=
proy ") M
M
= (5,5
M
= ( 25,15
3
= ( 5,−5 ()
3
")
M
3 ) − ( − 10,0 ) (+
3
)− 5,−5
3
+( 10 ) ( ,0 −) − 10,0
) − −) 5,−5
= (10,0)
3
M . ") 5 ") = 250 (10 = ) ( ,)0 = 2
=
− 5,)(5
proy ") M
100
")
2
10,0
1!.Dados los puntos P2?16!@B 82?!60@B R2?06;@B S2?<61>@. ,allar los puntos M ' N6 si se sae #ue:
=
M*
PS
( PS − QR ) 3
= ( -,10) ( −) 1(,2) =
8,8
QR = ( 5,8) ( − )2,(5 )= 3,3
REMP(A=AMOS:
( PS − QR ) M*
=
3
( ) ( 8),8 −
⇒ M* =
3,3
3
EN9ON)ES: QS
= ( -,10) ( −) 2(,5) =
PR = ( 5,8 ) ( −) 1,(2
)=
7,5
4,6
M*
= * − M = ( Q + QS )( − P) + r PR
M*
= P − Q + t QS − r PR
5 , 5 = (1,2) ( − ) 2( ,5 )+ t( 7,)5 − r 3 3 5 , 5 − (1,)3 =( ) t 7,(5) − r 3 3
4,6
4,6
( )
⇒ M* =
5,5 3
5 5 = 3 , 3
8 , 14 = 7t − 4r ,5t − 6r 3 3 #
8 3
= 7t − 4r
#
14 + 18r − 12r 15
8 = 21
8=
8=
r
-8 + 126r
− 12r
5
14 3
= 5t − 6r
14 + 18r
14 + 18r 15 14 + 18r
-8 + 60r 5
15
= 2-
t
30
REEMP(A=AMOS: M
= ( P + r PR )
M
2= (1,)2 + ( )
M
= , 15
*
= Q + $ # QS
*
157 = ( 2),5 + ( )
M
= ( P + r PR )
M
= (1,)2 + ( )
30
4,6
73 3- 5
75
2-
30
7 .5
4,6
= 157 75
= 15t =t 30
=t
73 3= , 15 5 * = Q + $ # QS 124- , 232 * = 75 15
M
1+. En el tri-n$ulo A7) se tiene: 3(C
(C
(!
= "(
=
"( 3
,allar S ' 9. Si
(! = S"! + t!C
= !" + "(
(! = !" + 3(C (! = !" + 3( C! + !( ) (!
=
!" 3!C − 4 4
REMP(A=AMOS: (! = S"! + t!C !" 4
−
3!C 4
= S"! + t!C
1 − S − !C 3 + $ = 0 4 4
!"
#
1 4
−S =0
1 "=− 4
3 4
+$ = 0 3 T
=−4
+ =
1/.En el paralelo$ra"o A7)D6 Si ()
= m " + n "!
" = ") + )
1 3
"
3 ")
= ") + )
= ( + ) () = 5 !( − ) () = 5 !( − 2 ")
2 ")
= )
()
= 5!( − 2
" ()
= 5( ! − ( ) − 2
()
= 5( !" + " − () − ) ) − 2 "
()
= 5!" + 5 " − 5 () − 5) − 2 "
3
3
3
"
6 ")
= 5!" + 15
6 ")
= 5!" + 13 " − 2 "
6 ")
= 5!" + 11
")
3
3
=
5 6
!" +
11 18
" 3 "
− )
3
( = 5!(
. Si
.
()
e mpla+amos :
6
" 3
! = !(
+ ( ! − ( = !(
=
")
−5 6
"! +
11 18
"
Re"plaa"os 11 18
" −
11
m=
5 6
"! = m " + n "!
∧n = −
18
m.n =
5 6
11
.
−5
18 6
=
− 55 18
10.En el paralelo$ra"o A7)D6 M ' N son puntos "edios. ,allar !sC+t. Si se sae #ue: C
= sMC + t*
C
= sMC + t*
C
= * + *C
C
= * +
!C
!C 2
= !M + MC
!C
=
!C
=
!C
=
!" + MC 2 C + MC 2 C + 2 MC 2
e mpla+amos :
C + 2MC 4
C
= * +
C
= − * −
5 C 4
MC 2
=
C
=
4MC
C
=
2 MC
10
C 4
+
MC 2
− * − 4 * −
5
4
*
5
e mpla+amos :
2MC 5
S
− 4 * = sMC + t* 5
2
−4
5
5
= ∧t =
Hallar
2 − 3 − 3 = 16 5 4 5
2
16.
E ncontrar el vector
´ dela "igura AB
y
b
8 12
p
15
o
x
Solucion :
´ =( 35,12 ) #$ u#$ ´´ =
´´ = u#$
35,12
√ 352 + 122
(
35 12
,
37 37
−12
⊥
) 35
u#$ ´´ = 37 , 37
(
(
´ = AB
381 600 37
,
37
→
13.Dados →
a
)
)
→
= (8,6) ∧ b = (−2,6)
→ → →
p ⊥ b , q // b
→
'
a
→
→
→
6 allar los %ectores
p q
'
→
= p+ q Solucin
→
→
p⊥b
→
→
q// b
tales #ue
→
→
→
p+ b
→
→
p+ b
→
= p− b
( q1 , q 2 ) = r (−2,6)
= ( p1 − 2* p 2 + 6)
( q1 , q 2 ) = ( −2r ,6r )
( p1 −) 2( 2 +) p 2 ( + 6 )2 (= ) p1 + 2 2 + p 2 − 6 1
q1
=
p1
= 3 p2
p1
=-
→
a
q1
= −2r ∧ q 2 = 6r
∧ q2 = 3 − 4 p + 12 p = 4 p − 12 p 1 2 1 2
2
→
2
→
= p+ q
(8,6) = (3 p 2 , p 2 ) + ( −2r ,6r ) 8 = 3 p2
6 = p2
= 10 ∧
20r
r
− 2r ∧
1
=
2
+ 6r
p2
= 6 − 6r
p2
=3
→
∴ p = (-,3) →
q
= (−1,3)
a
=
1;.Sea #
( a)1 2 ( + ) a 2
a+b
65
6 2
=
65
=
Comp a ( a b )
164
6
102
= a
.,allar la
Compb ( a − b )
( a)1 2 ( +) a 2 #
2
= 65
( a1 +) b(1 2 + ) a 2 + b2
( a1 +) b(1 2 + ) a 2 + b2
2
2
=
= 164
( )a1 2 + 2a1(b) 1 ( +) b1 2 + a( 2) 65 + 2a1b1 2( a1b1
+ ( b) 1 ( 2) +
b
a2
+ a 2b) 2( ) +( )b1 2 +
( b)1 2 (+ ) b2
2
164
2
+ 2 a 2 b2 +
2
+ (2)a 2 b2 +
b2
2
b2
b2 2
2
= 164
= 164
= --
= 25
= ( b)1 2 (+ ) b2
(
Comp b ( a − b ) =
2
=
25
a −)(b) (b
b
=5
)( a.)b =
b .b 5
=
37 − 5 5
=
32 5
1<. ,allar el -n$ulo &or"ado por los %ectores b
=
(1,3)
Cosθ
(a # b ) = a∗b
Cosθ
=
Cosθ
( ) = 1+ 6 =
(1,2)( 1,)3 5 # 10
50
7 5 2
a
'
r oyb ⊥ a
a
6 si
= (1,2)
'
θ
7 = arcCos 5 2
θ
= 8.13
!>.(os %ectores
a
'
b
a
de lon$itudes !'+ respecti%a"ente
)?161@ )alcular: r oy c ( a + b )
= r oy c ( a ) + r oyc ( b )
Donde:
) oy c ( a r
a + c = 2 c c a.c
)
(
r oy c a
b + c = 2 c c b .c
) Cos ( α = ac
)
(
Cos β
=
a.c
)
(
Cos β
=
) Cos( α =
2 2
2 2) Cos ( α
= a.c
r oy c ( a + b )
)
= r oyc ( a ) + r oy c ( b )
a.c b .c 2 c + 2 c c c 2
(
2Cos ( α ) (1,1) + 3 2Cos ( β ) (1,1) 2 2
2Cos ( α ) )(1) ,1
+ (3 2()Cos β )(1,1)
(
bc
b .c 3 2
3 2Cos β
= b .c
= 2∧ b = 3 6
!1.,allar el %alor de
b
r oy b⊥ ( a)
Compb a es %ositio
Compb a > 0, θ a,udo
= r oy b ()a + r oy () b⊥
rb
a
= ( − 4,)2( − ) − 3,3 = ( − 1,−1)
1 1 b
a
6
= ( − 4,2)
r oy b⊥ ( a ) // b
( − 4) ,2 =( rb) + − 3,3
rb
2
( = −) 3,3
r oyb⊥ ( a ) a
=2
b
saiendo #ue:
= rb =2
b
2
2
2
1 + 1 = 2 r r 1 r
r
2
+
1 r
=±
2
=2
2
2
1 2
= − r ,− r ⇒ b = ( − 2,−2)
22.
´ ‖=a%sean E´ y F ´ = AD ´ ,‖AB ´ !untos edos de loslados BC ´ y s ean A , B , C , Dunre ctangulo tal que 2 AB
´ = AE ´ + AC ´ + AF ´ , 'allar el valor de : si & ´ &
2 ´&
Co! AB ´ + Co! AD ´ .
B
E
C a/2 F a/2 D
A 2a
´ = AD ´ y‖ AB ´ ‖=a 2 AB ´ + AC ´ + AF ´ & = AE ´ &
´ 2&
co! AB ´ + co! AD ´ . Solucion :
´ AB ´ & co! = AB ´ ´ &
AB ´
´ &
co! AB ´ =
´ AB ´ & a
´ 2&
´ &
co! AD ´ =2 co! AD ´ =
´ &
´ 2&
´ AD ´ ) 2( &
(co! AB ´ + co! AD ´ =
´ AD
´ AB ´ 3&
´ + AC ´ + AF ´ & = AE u´ AD ´ = ( cosθ,senθ )
´ AD u´ AD ´ = ( 1,0 ) ∧ u ´ =( 0,1) ⊥
´ =2 a ( 1,0 ) + AF
a 2
( 0,1 )
a
=
´ 2 AB ´ ) 2 (& 2a
=2
´ AB ´ ) (& a
( )
´ = 2a, AF
a
2
´ =2 a ( 1,0 ) + a ( 0,1 ) AC ´ =( 2 a , a ) AC
´ =a ( 1,0 ) + a ( 0,1 ) AE ´ =( a , a ) AE
´ =CD ´ =a ( 0,1 ) AB ´ =( 0, a ) AB
´ =( a , a )+ ( 2 a , a ) + 2 a , a &
( ) 2
(
´ = 5a, &
⇒
5a
´ AB ´ 3& a ´ &
2
)
( =
3 5 a,
´ 2&
co! AB ´ + co! AD ´ =
5a 2
)(
a 15 a 2
0, a )
23.
´ = 4 FB ´ , E n el rectangulode la"ig ura ' , ) y $ son !untos edios = AB
´ = 4 a , #A ´ = a,si * ´ = &F ´ + A) ´ + $C ´ #C 'allar : *´
*´
co! AB ´ + co!$B ´ .
a/2
a/2
2a
2a
Solucion : u´ #$ ´ = ( cos0, sen 0 ) = ( 1,0 )
= ( 0,1 ) u´ #$ ´ ⊥
´ =2 a ( 1,0 ) + a ( 0,1 ) $B ´ =( 2 a , a ) $B
´ =2 a ( 1,0 ) $c ´ = (2 a , 0 ) $c ´ ⟧ =4 a Si ⟦ AB
⟶
´ =a FB
´ = a u´ #$´ + a u´ #$´ ⊥ 'F 2
( )
´ = a, 'F ´ = A)
a
2
´ 'F
´ − 'A
2
(
´ = A)
a 2
,−
a 4
( )+(
*´ = a ,
*´ =
(
a
a
2
2
7a a 2
)
,
4
,−
a 4
)+ (
2 a , 0)
)
´ =( 4 a , 0 ) AB
´ *´ $B ´ *´ AB + ´ ´ AB $B
(
7a a 2
,
4
)(
4a,0)
4a 7a 2
+
( +
7a a 2
,
4
)(
2 a ,0 )
a √5
29 a 4 √5
24.
E n el triangulo equilatero ABC de la "ig . & y + trisecan el segento de B a C
´ = A& ´ ,$ ´ = A+ ´ + AB, ´ Calcular si B ´ )
´ $
co! AB ´ + co! AC ´
B
c
M
b
N a
A
Solucion : deduciendo que a=
b=
4√3 3 8√3 3
c= 4 √ 3
(
´ = A+
20 4 3
,
3
√3
)
´ =4 (1,0 )+ 4 √ 3 ( 0,1 ) AB ´ =( 4,4 √ 3 ) . AB
C
´= $
(
20 4
´= $
(
32 16
,
3
,
3
)
√3 + ( 4,4 √ 3 )
3
3
√3
)
´ = A&
16 (1,0 )+ 8 3 3
´ = A&
(
16 8
,
3
´ )
Co! AB ´ =
3
´ )
AB
´ $ ´ AC
co! =
´ $
co! AC ´ = ´ )
)
√3
( =
16 8
( =
32 6
´ ´ AB !
´ ‖ ‖ AB
Co! ´ =
√ 3 ( 0,1 )
,
3
)
√ 3 ( 4,4 √ 3 ) 8
20 3
´ AC ´ $
3
,
´ ‖ ‖ AC
3
)
√3 ( 8,0 ) 8
32 3 ´ $
Co! ´ + co! AC ´ = AB
3
52 3
.
SECCIONES CÓNICAS
2
1. Hallar la ec ac!"# $e la c r%a
+ 2 − 2 − 4 − 4 = 0
&! &e real!)a #a 'ra&lac!"# $e e*e& al p#'o Solc!"#+ Co(ple'a(o& ca$ra$o& ( − 1) 2
e# el &!&'e(a
% 0 (1,2)
+ ( − 2) 2 = -..............(1)
,a relac!"# -e ay e#'re el &!&'e(a or!/!#al y el #e%o &!&'e(a %!e#e $a$o por
= 0 + ⇒ = − 1 = + = − 2 0 0e(pla)a#$o e# 1 C:
2
+ 2 = -
2. Hallar la ecac!"# $e la cr%a
2
− 8 − 8 − 8 = 0
&e real!)a #a 'ra&lac!"# $e e*e& al p#'o Solc!"#+ Co(ple'a(o& ca$ra$o& ( − 4) 2
= 8( + 3)
e# el &!&'e(a
&!
% 0 (−3,4)
333333I
Tra#&4or(a#$o = + = + 3
= + ⇒ = − 4 0 0
0e(pla)a#$o e# I 2 ) : = 8
. S! #a par6bola co# e*e 4ocal %er'!cal '!e#e # 4oco e# rec'o '!e#e #a lo#/!'$ $e $!rec'r!). Solc!"#+
12u
( 0 , 4)
y & la$o
7 allar & ecac!"#7 & 4oco y & rec'a
=
4%
= 12 ⇒ % = 3
Co(o el e*e 4ocal e& %er'!cal e#'o#ce& 2
2
= 4% = 12( − 1)
. %ar!ola
,a rec'a $!rec'r!) = −2
. Hallar la ecac!"# $e la c!rc#4ere#c!a -e pa&a por el %9r'!ce y lo& ex're(o& $el la$o rec'o $e la par6bola Solc!"#+ Se &abe -e
2
=
4
4% = 4 ⇒ % = 1
El %ec'or #!'ar!o -e %a e# $!recc!"# $el e*e 4ocal →
u
= (1,0)
u
⊥
→
→
C = − C = % u →
= (0,1) ⇒ = (1,0)
= − = 2%u ⊥ ⇒ = (1,2)
→
C+
=
:a(o& a &po#er -e →
C
=1
→
*
C
=
→
5
*
Se/;# el 'r!6#/lo F0A
2
+ + 2 = + 2 .......(1)
Se/;# el 'r!6#/lo C0A
→
C +
+ =2
e& el $!6(e'ro $e la par6bola
5 + +
2
= (1 + + ) 2 .......(2)
De la ecac!"# 1 y 2 +
= 4 ⇒ C+ = 5
,a ecac!"# $e la c!rc#4ere#c!a &er!a 2 C:
+ 2 = 5
7 5. Hallar la ecac!"# %ec'or!al y car'e&!a#a $e la par6bola -e '!e#e por ( 0,2) −−2=0 4oco al p#'o y por $!rec'r!) a la rec'a allar & %9r'!ce y lo& ex're(o& $e & la$o rec'o. Solc!"#+ De la rec'a $!rec'r!) 'e#e(o& →
→
= (1,) −1 ⇒ u (=)
d[ , ] 2
=
= 2% =
1 2
1,−1
→⊥
*
u
=
(1,1) 2
+ + + C +2
1(0) − 1(2) − 2 2
+ 2 ⇒=
2
,a2 ecac!"# %ec'or!al $e la par6bola e&
2
= 4% = 4 2
Do#$e
= [ ( , ) − C] →u ⊥ = [ ( , ) − C] →u
→
→
9 = − % u
Sea →
⇒ 9 = ( − 1,3) el la$o rec'o $e la par6bola
= 4% ⇒ = 2 = −=
⇒ = ( 2,4)
→ →⊥
u
2
=2
2
(1,1) 2
= ( 2,2 )
= (−2,4)
<. =#a c!rc#4ere#c!a '!e#e & ce#'ro e# el 4oco $e la par6bola $e ecac!"# 2
)
− 12 − 36 = 0
+ y pa&a por el %9r'!ce $e e&'6. Hallar & ecac!"#. Solc!"#+ Se &abe -e #a par6bola '!e#e $e ecac!"# 2 4%
⇒ 2 = 12( + 3) 4% = 12 ⇒ % = 3 d[ 9, ] [=] d 9, = = ) Co(o 0>ra$!o $e la c!rc#4ere#c!a
C
+
2
+ 2 = -
?. ,o& ex're(o& $el la$o rec'o $e #a par6bola
)
&o# @712 y ?7B y
→
9
la& co(po#e#'e& $el %ec'or :>%9r'!ce7 F>4oco &o# po&!'!%a&. E#co#'rar la& ecac!o#e& %ec'or!ale& $e la par6bola ) y & $!rec'r!) l
Solc!"#+
+ ,a$o rec'o
→
= ( − 16,12 ) ⇒ u
→
= ( − 3,−) 4 /( 5) =
u
= 4 =
→
→
→ →
=
( − 4,3) 5
→⊥
=u
3,4 / 5
= 20 ⇒ = 5
→
9. = u 9
(
= −
4,2 )
,a ecac!"# %ec'or!al $e la par6bola e& 2 = 4%
): →
2 = 20% →
'9 = u
*
'∈
' = ( − 7,−2 ) →
: = 0 + t a * t ∈ ℜ : % = (−7,−2) + t ( − 4,3)
8. 2 Hallar la ecac!"# $e la el!p&e co# %9r'!ce& e=
exce#'r!c!$a$ $!rec'r!ce&.
/ −1.−1 -
y
/ 3.3 -
y
1 2
a$e(6& allar lo& 4oco&7 el ce#'ro y la& rec'a&
SOLUCIÓN &1
= / −1.−1 -
&2
= / −1.−1 -
e=
1
u=
/ 1.1 -
u
⊥
2
2
=
/ 1.1 2
Se &abe -e la $!&'a#c!a $e lo& %9r'!ce& e&+ 2a
= &1&2 = &2 − &1 = / 4 .4 -
→ a = 1/ 2
e=
c a
42
+ 42 - = 2
→ c = 1/2
2-=
2
2 2
E# la el!p&e &e c(ple la &!/!e#'e relac!"# p!'a/"r!ca+
= a 2 − b2 2 = 8 − b2 → b =
c2
x0 2 a2
−
y0 2 b2
=1
6
7 $o#$e+
x 0 = [/ x . y - − C ]u y 0 = [/ x . y - − C ]u ⊥ x 0 = [ x − 1. y − 1] y 0 = [ x − 1. y − 1]
/ 1.1 -
=
2 / −1.1 2
x+ y−2 2
=
− x +1+ y −1
0e(pla)a#$o ( x + y −) 2( 2 ) 16
+
y−x 12
2
=1
,a ecac!"# $e la el!p&e e&
2
=
y−x 2
ε 1 7 x 2 + 7 y 2 − 2 xy − 12 x − 12 y − 36 = 0 →
→
C)1
= C)1 u / 1.1 -
)21
−C =
)1
= / 1.1 - + / 1.1 - = / 2 .2 -
→
)2 C
=
C − )22 )2
2
2
→
)2 C u
=
2
/ 1.1 2
= / 1.1 - − / 1.1 - = / 0.0 -
→
→ →
CQ = CQ u
Q −C
=4
/ 1.1 -
= / 2.2 2 → Q = / 1.1 - + / 4.4 - = / 5.5 23
se tiene: →
L1 1 P = Q + t u
P = / 5.5 - + t/ −1.1 →
Q0 C
→
→
= Q0 C u
C − Q0 = 4 2
/ 1.1 -
2 Q 0 = / 1.1 - − / 4 .4 - = / −3.−3 -
⇒ L2 1 P = / −3.−3 - + +/ −1.1 -
1. 25Hallar la ecac!"# $e #a el!p &e cyo& 4oco& &e e#ce#'ra# e# la !#'er&ecc!"# $e la& rec'a& e=
y & exce#'r!c!$a$ e&
L1 1 2 x − y
=1
L2 1 3x − y
= 14
3 5
7 allar & ce#'ro y && %9r'!ce&.
L1x−y
=0
Solución
)1 ∈ L1 L )2 ∈ L2 L L1 1 2 x − y
=1
7 e=
L2 1 3x − y 3
5 L1x−y =0 )1 ∈ L1 L •
2 x − y = 1 x − y = 0 x =1 . y =1 )2 ∈ L2 L •
3x − y = 14 x − y = 0 x=7 . y=7
− 14
→
= / 6.6 -
)1 )2 →
b
=
/ 1.1 2
=
2c
e=
)2
→a =5
•
→
=
C&2 &2
− )1 = 6
2
c a 2 →
C&2 u
=C +5
2
/ 1.1 2
= / -.- &1 = / 4 .4 - − / 5.5 &1 = / −1.−1 &2
•
2. 2<,a c!rc#4ere#c!a
C 1 / x − 3 -2
+ / y + 2 -2 = 100
e&'6 c!rc#&cr!'a a #a ( -,6) ∈ ε ε e = 1/ 2 el!p&e $e exce#'r!c!$a$ -e pa&a por .Hallar la ecac!"# %ec'or!al $e la el!p&e y la& ecac!o#e& $e la& $!rec'r!ce& $e la el!p&e. SOLUCIÓN DATOS+ c 1 / x − 3 -2 e=
1 2
+ / y + 2 -2 = 100
P0
= / -.6 - ∈ ε
i. •
ε
La ecuación vectorial de la elipse E#co#'ra(o& la $!&'a#c!a $el ce#'ro a&'a el :17 -e e& el &e(!e*e 'ra#&%er&al+ a = C&1 4 ( C − &1 ) = / 6.8 -
=
a
62
+ 8 2 = 10
Se puede encontrar el vector unitario 1 →
u •
/ 6 .8 10
- 4
= / 5 .5 - =
/ 3.4 5
Se &abe -e la exce#'r!c!$a$ $e #a el!p&e e&+ e=
•
= c a
→c=
1 2
/ 10 - = 5 2
Se c(ple la &!/!e#'e relac!"# p!'a/"r!ca + c2
= a 2 − b2
50 = 100 − b 2
,a ecac!"# $e #a el!p&e e# el #e%o &!&'e(a e& $e e&'a 4or(a
• 2
x0 a2
+
y0 2 b2
=1 2
⇒ ε 1 x0 + 100
2
y0 50
=1
!!. Las ecuaciones de las directrices de la elipse. • E#co#'ra(o& la $!&'a#c!a -e ay $e&$e el ce#'ro a&'a el p#'o +
→
→
→
CQ = CQ u
. Q ∈ L1
/ 3.4 5 Q = / 3.−2 - + / 6 2 .8 2 Q =C
+ 10
23
Q = / 3 + 6 2 .8 2
− 2-
⇒ L1 = p = / 3 + 6
2 .8 2
− 2 - + t/ −4 .3 -
. 2? Hallar la ecac!"# $e la H!p9rbola y && ele(e#'o& &! & ce#'ro e& / −3.−1 -
%9r'!ce 17@1 y 4oco SOLUCIÓN
c = ae → e = •
→
b = C)
c a
=
/ 2 .−1 -
17 15
= / 0.8 - → b = 8
• •
Halla(o& el %ec'or #!'ar!o+
.
→
u
→
=
C)
=
→
C)
/ 0 .8 8
= / 0.1 -
⇒ u ⊥ / −1.0 •
Se &abe -e la $!&'a#c!a $el ce#'ro a&'a el 4oco e& !/al a c+ →
C) •
02 + 17 2
= a 2 + b2 → a =
17 2
− 8 2 = 15
.
donde 1
+ ,a ecac!"# $e la !p9rbola %!e#e $a$a $e la &!/!e#'e 4or(a+ y0 2 b2 ⇒
−
x0 2
=1
a2
y0 2
−
64
x0 2
=1
225
x 0 = [/ x . y - − C ] u ⊥
y 0 = [/ x . y - − C ] u
•
= 17
or c(pl!r&e la& &!/!e#'e relac!"# p!'a/"r!ca7 'e#e(o& c2
•
= c → c = / 0.17 - =
Halla(o& la rec'a , 1+ L1 1 Q ∈ L1 →
CQ
=a= e
∴Q = C +
→
→
→ CQ = CQ u 15 17 5 5
15 2 17
= 15
2
17
3/ 0.1 -
Q = / 0 .2 - + / 0 .
15 2 17
-
Q = / 0 . 1532 L1 1 P = / 0 . 1532 - + t/ −1.0 -
e=
'
. B7 S ea rec'a& Q =/
la !p9rbola cya exce#'r!c!$a$ e& a&#'o'a&
l1 1 11x − 2 y − 25 = 0
&o#
157 6 . -∈ ' 5 5
l2
.Hallar+
Solución
i.
'
La ecuación de
'
Halla(o& el ce#'ro C(/ l1 l2 - = /
P =/
'
+
5 →
•
Co(o →
u n1 •
11
11 5
.
.
−2
−2
5
⊥
- + x0 2 u + y0 2 u 5
→
x0 2 2
a
5
u 55 x 0 → u
→
→
= u a + ub
= / 12 .−4 - = 4/ 3.−1 = / 11.−2 -
-
+
y0 2 2
b
=1
5 2
y cya&
= x − 2y − 3 = 0
.
S!
→ = / 2 .11 - → u →a = 2 . 11 5 5 5 5
→
a
n2
= / 1.−2 -
→ = / 2 .1 - → u →b =
→
b
→
•
⊥
5
.
1
5
12
⇒ u =/5 u
2
=/−
16 3 4 5 . 5 5 - = t/ 4 . 5 -
4 3 . 5 5
Tra#&4or(a#$o al &!&'e(a
x0 y 0
→
x 0 = [C − Q] u
x 0 = / −
11 2 17 6 3 4 . - + / . ./ -=2 5 5 5 5 5 5
y 0 = [C − Q] u
⊥
6 8 4 3 − 24 + 24 = 0 y 0 = . − . = 5 5 5 5 25 25 '
Q 0 = ( 2 .0) ∈ •
4 a2
=1→ a = 2 ∧ c =
5
Se c(ple la relac!"# p!'a/"r!ca+ c2
= a 2 + b2 ⇒ b = 1
'
ii.
⇒
⇒
x0 2 4
− y 0 2 = 1 4 y0 2 − x0 = 1 4
Las ecuaciones cartesianas de las bisectrices a ⊥ u - + tu e
L1 1 P = / C + L1 1 P = //
11 5
.
−2 5
-+/
12 16 −4 3 . -- + t/ . 5 5 5 5 5 5
L2 1 P
= /C −
L2 1 P = //
11 5
a ⊥ u - + tu e
.
−2 5
-−/
−4 3 12 16 . -- + t/ . 5 5 5 5 5 5
