dx/dt = 1
dy/dt = -2t
hence dy/dx = dy/dt divide by dx/dt = -2t
Hence, gradient of tangent at any point = -2t
Which means that
gradient of normal at any point = 1/2t since the 2 gradients must
multiply to -1
We can understand by at the point
(2+t, 1-t^2), gradient of normal = 1/2t
Hence, using equation of straight line y = mx + c
y = 1/2t x + c
to find c, sub point (2+t, 1-t^2)
1-t^2 = 1/2t (2+t) + c
c= 1/2 - t^2 - 1/t
Thus,
y = 1/2t x + 1/2 - t^2 - 1/t
Multiply by 2t throughout
2ty = x + t - 2t^3 - 2
Rearranging gives you
x - 2ty = 2t^3 -t + 2
