Question from http://www.sgforums.com/forums/2297/topics/272230
A curve C is given parametrically by the equations:
x = 2 + t , y = 1 - t²
Show that the normal at the point with parameter t has equation:
x - 2ty = 2t³ - t + 2
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Answer:
dx/dt = 1
dy/dt = -2t
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-t²), 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-t²)
1 - t² = 1/(2t) * (2+t) + c
c= ½ - t² - 1/t
Thus,
y = 1/(2t) * x + ½ - t² - 1/t
Multiply by 2t throughout
2ty = x + t - 2t³ - 2
Rearranging gives you
x - 2ty = 2t³ -t + 2
