２０分!amaths1問

x=acos^3 t , y=asin^3 t

first we have to find the slope of the tangent line.

To find this, we find dy/dx

dy/dx=dy/dt ． dt/dx (Chain rule)

where dy/dt = 3asin^2 tcos t , dx/dt=3a cos^2 t (-sin t)

i.e. dy/dx =3asin^2 tcos t / 3a cos^2 t (-sin t) = -tan t

the slope at the point thus is -tan pie/4=-1

when t = pie/4, x =y= (開方2)a/4 (by direct substitution)

i.e. equation of the tangent is: y-(開方2)a/4=(-1)(x- (開方2)a/4 )

x+y=(開方2)a/2

x+y=a/(開方2)

x=acos^3 t , y=asin^3 t
when t =π/4, x = a * (1/(2*開方2)), y = a * (1/(2*開方2))
dy/dt =3a [sin t]^2
dx/dt =3a [cos t]^2

dy/dx = [tan t]^2
when t = π/4 , dy/dx = 1
=> 切線 slope = -1

consider slope of line
-1 = (y - a/(2*開方2))/ (x - a */(2*開方2)))
-x + a/(2*開方2) = y - a/(2*開方2)
x + y = a/(開方2)

dy/dt=a(3sin^2 t)(cos t)
dx/dt=a(3cos^2 t)(-sin t)
dy/dx=-sin t/cos t
dy/dx=-tan t
切綫斜率=-tan pie/4 =-1
切點的x座標=acos^3 (pie/4)=a開方2 /4
切點的y座標=asin^3 (pie/4)=a開方2 /4
方程:
y-a開方2 /4=-(x-a開方2 /4)
y-a開方2 /4=-x+a開方2 /4
x+y=a開方2 /2 <==a開方2 /2=a/開方2<==分母有理化咗

dx/dt = a[3cos^2 t (-sin t)] = -3acos^2 t sin t
dy/dt = a(3sin^2 t (cos t)] = 3asin^2 t cos t
dy/dx = (dy/dt)/(dx/dt) = -sint/cost
切線斜率 = dy/dx at π/4 = -sin(π/4) /cos(π/4 ) = -1
當t = π/4 , x = a cos^3 (π/4 ) = a/2sqrt(2)
y = asin^3 (π/4 ) = a/2sqrt(2)
切線方程:
y - a/2sqrt(2) = -(x-a/2sqrt(2))
2sqrt(2) y - a = -2sqrt(2) x + a
2sqrt(2) x + 2sqrt(2) y - 2a = 0
x + y - a/sqrt(2) = 0
x + y = a/sqrt(2)