關於Differentiation的簡單A.Maths.問題

A curve is represented by the parametric equations
x = 4 / (2 + t)2, y = 12 / (2 + t)
(a) Find the equation of the chord joining the points P and Q with parameters t = -1 and t = 0 respectively.
t = -1 => x = 4; y = 12
t = 0 => x = 1; y = 6
Equation of the chord = (y – 6)/(x – 1) = (12 – 6)/(4 – 1) = 2
y – 6 = 2x – 2
y = 2x + 4
(b) Find dy / dx in terms of t.
x = 4(2 + t)-2
dx/dt = 4(-2)(2 + t)-3 = -8(2 + t)-3
y = 12(2 + t)-1
dy/dt = 12(-1)(2 + t)-2 = -12(2 + t)-2
dy/dx = (dy/dt) / (dx/dt) = [-12(2 + t)-2] / [-8(2 + t)-3]
dy/dx = (3/2)(2 + t)
(c) Find the equation of the normal to the curve which is perpendicular to PQ.
The normal is perpendicular to PQ => the tangent is parallel to PQ
Slope of PQ is 2, so (3/2)(2 + t) = 2
2 + t = 4/3
t = -2/3 =>
x = 4 / (2 – 2/3)2 = 9/4;
y = 12 / (2 – 2/3) = 9
Equation of normal = (y – 9) / (x – 9/4) = -1/2
2y – 18 = 9/4 – x
y = 81/8 – x/2