Kakšna je standardna oblika enačbe parabole s fokusom na (1,7) in direktorom y = -4?

Odgovor:

# y = x ^ 2/22-x / 11 + 17/11 # standard od

# (x-1) ^ 2 = 22 (y-3/2) # Vertexna oblika

Pojasnilo:

iz dane Focus #(1,7)# in directrix # y = -4 # compute # p # in vertex # (h, k) #

# p = (7–4) / 2 = 11/2 #

vertex # h = 1 # in # k = (7 + (- 4)) / 2 = 3/2 #

vertex # (h, k) = (1, 3/2) #

uporabite obliko vozlišča

# (x-h) ^ 2 = 4p (y-k) #

# (x-1) ^ 2 = 4 * 11/2 (y-3/2) #

# (x ^ 2-2x + 1) = 22 (y-3/2) #

# x ^ 2-2x + 1 = 22y-33 #

# x ^ 2-2x + 34 = 22y #

# (x ^ 2-2x + 34) / 22 = (22y) / 22 #

# (x ^ 2-2x + 34) / 22 = (preklic 22y) / preklic22 #

# y = x ^ 2/22-x / 11 + 17/11 # standard od

graf {(y-x ^ 2/22 + x / 11-17 / 11) (y + 4) = 0 -20, 20, -10, 10}