CALCULO DE CONICAS

8/19/2019 CALCULO DE CONICAS

1/37

construir la parabola

vertice (3,2) se encuentran los puntos en el eje x

foco (3,4)

lado recto p4 8 p= (4-2) lado recto en

directriz vertice-p 0

formula vertice (3,2)

x

vertice 3 2

foco 3 4

-4 8!"2#-3 $!#

-2 #!"2#

-" 4

0 3!"2#

" 2!#

2 2!"2#

3 2

4 2!"2#

# 2!#

$ 3!"2#% 4

8 #!"2#

& $!#

"0 8!"2#

lado recto en los puntos foco (3,4) (-",4)

ecuacion de la recta ∆y

m= ∆y '2-'" ∆x

∆x 2-"(4-4) 0

(-"-3) -4

y=4

*+ ./

'

(x-1)2=4p(-) (x-3)2=4p(-2)

(x-3)2=4(2)(-2)

(x-3)2=8(-2)

x2-$x&=8-"$

x2-$x-82#=0

8=x2-$x2#

x2-$x-82#=0

8=x2-$x2#

diferencia dabs

y-y1=m(x-x

1) y-4=0(x-3)


2/37

4 -"

4 0

4 "

4 2

4 3

4 4

4 #4 $

4 %


3/37

el foco x=-b2a 3

=&-"82#8 2

0

8=x2-$x2#

ordenada isa

-# -4 -3 -2 -" 0 " 2 3 4 # $ % 8 &

0

"

2

3

4

#$

%

8

&


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"0 ""


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1ip6rbola si x es - vertical si x es es 1orizontal

formula

vertical 1orizontal

- -

obtener raiz

para obtener c

centro

/=(1,) -1= -2 por si7nos 1= 2

-= -" = "

focos

vertices

asintotas

=- (ba)(x-1)

lado recto

exentricidad

e=ca

ecuacion de la recta ∆y

m= ∆y '2-'" ∆x∆x 2-"

ecuacion punto pendiente

(-)2 (x-1)2 (-)2 (x-1)2

a2 b2 b2 a2

a2= √a

b2= √b

c2= a2 b2

f "=(1,c)

f 2=(1,-c)

v"=(1,a)

v2=(1,-a)

r=2b2a

diferencia deordenada

abscisa

-"=m(x-x

")


8/37

-3 -2 -" 0 " 2 3 4 # $ %

-%

-$

-#

-4

-3

-2

-"

0

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2

3

4

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$

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"2

"3

"4

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asintota

asinto -

parametro

. +59: .; *


9/37

factor coom>n

completar los cuadradosdivir entre "2

vertical

-

2 ba "!%320#08"

3!4$4

/=(1,) (2,") -1= -2 2

(2,"4) (2,#)

(2,"-4) (2,3)

(2,"2) (2,3)

(2,"-2) (2,-")

= (ba)(x-1) -2!4$ para x=0

=- (ba)(x-1) 4!4$

"2 $e=ca 2

7ra?car asintotas x

centro= (2,") 2 "

asintota -2!4$ 0 -2!4$

ecuacion de la recta

m= ∆y '2-'" -3!4$

∆x 2-" -2 "!%3

ecuacion punto pendiente -"!%320#08"

-"="!%3(x-2) -"=("!%3x)-("!%3@2) =-("!%3x)(

asintota 2!4$4"0"$2

-2 %!&282032303

-" $!"&$"#2422%

0 4!4$4"0"$"#"

" 2!%320#080%$

2 " "

32-x2-$4x-"3

32-x2-$4x="3

32-$-x24x="3

(32-$)-(x24x)="3

3(2-2)-(x24x)="3

3(2

-2")-(x24x4)="33-43(2-2")-(x24x4)="2

3(2-2")-(x24x4)="2

(2-2")-(x24x4)="

(-)2 (x-1)2

(-")2-(x-2)2=" a2 b2

a2=Aa

b2=Ab

c2= a2 b2

f "=(1,c)

f 2=(1,-c)

v"=(1,a)

v2=(1,-a)

r=2b2a r=b2a

(x"-

")

-"=m(x-x

")

"2 "2 "2

4 "2

4 "2


10/37

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# -4!"&$"#2422%

$ -#!&282032303

7ra?car asintotas x

centro= (2,") 2 "asintota - 4!4$ 0 4!4$

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2!4$4"0"$2

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0 -2!4$4"0"$"#"

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2 "

3 2!%320#080%$

4 4!4$4"0"$"#"

# $!"&$"#2422%

$ %!&282032303

x

/=(1,) (2,") -1= 2 "

(2,"4) (2,#) 2 #

(2,"-4) (2,-3) 2 -3

(2,"2) (2,3) 2 3(2,"-2) (2,-") 2 -"

+59:* *


11/37

-= -" "

-2!4$4"0"$2

4!4$4"0"$2

"!%3@2)"


12/37

m= 3!4$

-2 -"!%3

"!%320#08"

"!%3@2)-"

4 0!33333333

4 0!33333333

4 0!333333334 0!33333333

4 0!33333333

4 0!33333333

4 0!33333333

4 0!33333333

4 0!33333333


13/37

(x-1)(-)

factorizar

completar cua drados

dividir por 4

2

"

/=(1,) (-4,-2) -1=4

(-42,-2) (-2,2)

(-4-2,-2) (-$,2)

-2!2$%&4&"&

-#!%320#08"

(-4,2") (-4,3)

(-4,2-") (-4,")

0!8$$02#4

"

Los elementos de la elipse son:

/=(1,) (-4,-2) -1=4

(-42,-2) (-2,2)

(-4-2,-2) (-$,2)

x2428x-"$28

x2428x-"$=-28

x28x42-"$=-28

(x28x)4(2-4)=-28

(x28x"$)4(2-44)=-28"$"$

(x28x"$)4(2-44)=-4

(x28x"$)4(2-44)=4

(x28x"$)(2-44)="

(x24)2(2-2)2="

a2

b2

a2=b2c2 √3

Vértices: V(h+a,k)Vértices: V'(h-a,k)Focos: F(h+c,k) (-4√3,2)Focos: F'(h-c,k) (-4-√3,2)Vértices del eje menor: B(h,k+b)Vértices del eje menor: B'(h,k-b)Excentricidad: c/a √32)

!: 2b2/a 2@"22

"entro: (h,k)Vértices: V(h+a,k), V'(h-a,k)Focos: F(h+c,k), F'(h-c,k)Vértices del eje menor: B(h,k+b) B'(h,k-b)Excentricidad: c/a!: 2b2/a

Vértices: V(h+a,k)Vértices: V'(h-a,k)

44

4 "

4 "

a2 b2


14/37

-2!2%

-#!%3

(-4,2") (-4,3)

(-4,2-") (-4,")

0!8$$02#4

"

Focos: F(h+c,k) (-4√3,2)Focos: F'(h-c,k) (-4-√3,2)Vértices del eje menor: B(h,k+b)Vértices del eje menor: B'(h,k-b)Excentricidad: c/a √32)

!: 2b2/a 2@"22


15/37

para x para

-=-2

x

-=-2 -4 2

-2 2

-$ 2

(x2) (2) (x2) (2)a2 b2 b2 a2

-% -$ -# -4


16/37

-2!2% 2

-#!%3 2

-4 3

-4 "


17/37

-3 -2 -"

0

0!#

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"!#

2

2!#

3

3!#

/olumn B


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5.. * ./9*/ +** ' C..* +. //9 D9. E*;* E

/. (1,) *+ r

ecuacion ordinaria

ecuacion 7eneral

y=mx#$ F%!&' E*+,E*E +E

E"'",%* +E '*% E*+,E*E +E !E"E"'",%* +E E*+,E*E dado dos .ntos

distancia entre dos puntos

obtener centro *($,3), 5(2,")

($2)2 4(3")2 2

("-3)= -2 "

(2-$)= -4 2

"E*!% (42)

y-=-(x-4) y-=-x# y=-x#10

obtener centro 5(2,"), /(4,&)

(24)2 3

("&)2 #(&-")= 8 4

(4-2)= 2 "

"E*!% (325)

y-5=-164(x-3) y-5=-x64#364 y=-x64#364#5

y=-x#10 2x="0 ecu "x#4y=3 x4=23 ecu 2

2x8=4$ ecu 3 multiplicamos @ 2 a ecu " para

2x8=4$ ecu 3

2x="0 ecu "%=3$ restamos

=3$%

x4(3$%)=23 sustituir en dos por tener la x factor

%x"44="$" multiplicamos @ %

%x="% despejamos

2("%%)="0 sustituimos en ecu "

(x-1)2(-)2=r2

x22-ax-bc=0

y-y1=m(x-x

1)

Fx=(x1#x

)6

Fy=(y1#y

)6

√(x-x

1)#(y

-y

1)

Fx=(x1#x)6Fy=(y

1#y

)6

y-y1=m(x-x

1)

Fx=(x1#x

)6

Fy=(y1#y)6

y-y1=m(x-x

1)

x"="%%

2-

"x2-

x"

m=

2-

"x2-

x"

m=

2-

"x2-

x"

m=


22/37

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/. (1,) ("%%,3$%)

radio 5(2,")

resolvemos lo int

radio= √(50647

resolvemos

acomodamos

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(x-)#(y#)

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(x-(-2)2(-(2)2=r2

*

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/. *!5 (4,2)

/. 5!/ (3,#)

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CALCULO DE CONICAS

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