Question

Tentukan persamaan garis singgung melalui
titik A(-5,0) pada kurva x - 4y² -1= 0


@ Wello1
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Answer 1

X - a = m(y - b)
x + 5 = my
m = (x + 5)/y

x - 4y^2 - 1 = 0
x = 4y^2 + 1
x' = 8y

x' = m
8y = (x + 5)/y
8y^2 = x + 5
y^2 = (x + 5)/8

x - 4y^2 - 1 = 0
x - 4(x + 5)/8 - 1 = 0
x - (1/2)x - 5/2 - 1 = 0
(1/2)x - 7/2 = 0
x - 7 = 0
x = 7


x - 4y^2 - 1 = 0
7 - 4y^2 - 1 = 0
4y^2 = 6
y^2 = 6/4
y = ±√(6/4)

m = ±8√(6/4)

x + 5 = my
Jadi, persamaan garisnya adalah
x + 5 = 8√(6/4)y atau x + 5 = -8√(6/4)y

Answer 2

    x-4y²-1 = 0
dx - 8y dy = 0
       dx/dy = 8y....... anggap titik singgung di (x,y)
     
m = Δx/Δy

(x+5)/y = 8y
     x+5 = 8y² ; subtitusi 8y²= 2x-2
     x+5 = 2x-2
         x = 7...........maka y = √(12/8)
                                      y = (1/2)√6 atau y = -(1/2)√6

titik singgung di (7, (1/2)√6):
m = Δy/Δx
m = ((1/2)√6)/12
    = (√6)/24
persamaan garis:
 y-0 =  ((√6)/24)(x+5)
24y = x.√6 + 5√6
     
titik singgung di (7, -(1/2)√6):
m = Δy/Δx
m = (-(1/2)√6)/12
    = (-√6)/24
persamaan garis:
 y-0 =  ((-√6)/24)(x+5)
24y = -x.√6 - 5√6