M2 Derivatives

Find the equation of tangents drawn fromthe origin to the curve y=4x^2－x+1
Sol
方法1
y=4x^2－x+1
y’=8x－1
8x－1=y/x
8x^2－x=y=4x^2－x+1
4x^2=1
x=+/-(1/2)
(1) x=1/2
y’=8*(1/2)－1=3
切線方程式：3x－y=0
(2) x=－1/2
y’=8*(－1/2)－1=－5
切線方程式：5x+y=0
方法2
設切點(m，n)
切線方程式：nx－my=0
nx=my
y=4x^2－x+1
my=4mx^2－mx+m
nx=4mx^2－mx+m
4mx^2－(m+n)x+m=0
D=(m+n)^2－16m^2=0
m^2+2mn+n^2－16m^2=0
n^2+2mn－15m^2=0
(n－3m)(n+5m)=0
n=3m or n=－5m
(1) n=3m
切線方程式：nx－my=0
3mx－my=0
3x－y=0
(2) n=－5m
切線方程式：nx－my=0
－5mx－my=0
5x+y=0

let y =mx be tangents
mx = 4x^2 - x +1
4^2 - (m+1)x +1 = 0
(m+1)^1 - 4(4)(1 )=0 (because tangent i.e. determinant = 0)
m^2 + 2m-15=0
m=3 or -5
so tangents are y=3x or y=-5x