Persamaan garis singgung parabola y= akar x + 1 ( hanya x yang diakar ) melalui titik (-8,0) adalah... Tolong dijawab ya soalnya buat dikumpul bsok.

Misal persamaan garis singgung tersebut :
y = mx + c
Melalui (-8, 0) => 0 = m(-8) + c => c = 8m
y = mx + 8m

y = y
mx + 8m = √x + 1
mx + 8m - 1 = √x ===> kedua ruas dikuadratkan
(mx + 8m - 1)^2 = (√x)^2
(mx + 8m - 1)(mx + 8m - 1) = x
m^2 x^2 + 8m^2 x - mx + 8m^2 x + 64m^2 - 8m - mx - 8m + 1 = x
m^2 x^2 + 16m^2 x - 2mx - x + 64m^2 - 16m + 1 = 0
m^2 x^2 + (16m^2 - 2m - 1)x + (64m^2 - 16m + 1) = 0
Menyinggung D = 0
b^2 - 4ac = 0
(16m^2 - 2m - 1)^2 - 4m^2(64m^2 - 16m + 1) = 0
(16m^2 - 2m - 1)(16m^2 - 2m - 1) - 256m^4 + 64m^3 - 4m^2 = 0
256m^4 - 32m^3 - 16m^2 - 32m^3 + 4m^2 + 2m - 16m^2 + 2m + 1 - 256m^4 + 64m^3 - 4m^2 = 0
-32m^2 + 4m + 1 = 0
32m^2 - 4m - 1 = 0
(4m - 1)(8m + 1) = 0
m = 1/4 atau m = -1/8

Jadi persamaan garis singgung :
y = mx + 8m

1) m = 1/4
y = 1/4 x + 8(1/4) ====> kali 4
4y = x + 8
x - 4y + 8 = 0

2) m = -1/8
y = -1/8 x + 8(-1/8) ===>
8y = -x - 8
x + 8y + 8 = 0