Question

es urgente por favor!!

Answer 1

I. La coordenada faltante del foco:  

1. (x+5)² = 20(y-10) ; F1 ( -5, ? )

Si, 4p = 20 ⇒ p = 5

h = -5  

k = 10

k+p = 10+5 = 15  

F1 (-5, 15 )

2. x² = 24y ; F2 ( 0, ? )

Si, 4p = 24 ⇒ p = 6

h = 0

k = 0

k+p = 0+6 = 6

F2 ( 0, 6)

3. (y+5)² = 60(x-10) ; F3 ( ?, -5 )

Si, 4p = 60 ⇒ p = 15

h = 10

k = -5

h+p = 10+15 = 25  

F3 (25, -5)

4. (y-15)² = 80(x-15) ; F4 ( ?, 15 )

Si, 4p = 80 ⇒ p = 20

h = 15

k = 15

h+p = 15+20 = 35  

F4 (35, 15)

II. La ecuación de la parábola correspondiente a cada V, F:

Sabiendo que,

(x-h)²=4p(y-k) siendo V(h, k) y F(h, k±p) ;

(y-k)²=4p(x-h) siendo V(h, k) y F(h±p, k) ;

• V(-2, 3) ; F(-2, 4)

Si, h = -2; k= 3 ; k+p=4

p = 4 - 3 = 1  

(x + 2)²= 4(y-3)

• V(-3, 1)  ; F(0, 1)

Si, h = -3; k= 1 ;  h-p=0

p = 3

(y-1)²= 12(x+3)

• V(1, 2)  ;  F(1, -1)

Si, h = 1 ; k = 2 ; k+p=-1

p= -1-2 = -3

(x - 1)²= -12(y-2)

• V(2, -3) ; F(2, -4)

Si, h = 2 ; k = -3 ; k-p=-4

p= -4+3 = -1  

(x - 2)²= -4(y+3)

• V(3, -1) ; F(0, -1)

Si, h = 3; k= -1 ;  h+p=0

p = -3  

(y+1)²= -12(x-3)

• V(-1, -2) ; F(-1, 1)

Si, h = -1 ; k = -2 ; k+p=1

p= 1+2 = 3

(x + 1)²= 12(y+2)