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The angle of intersection between the curve x2 = 32y and y2 = 4x at point (16, 8) is

Solution :

x2 = 32y    dydx = x16    y2 = 4x   dydx = 2y

  at  (16, 8), (dy\over dx)_1 = 1, (dy\over dx)_2 = 1\over 4

So, required angle = tan^{-1}({1 – {1\over 4}\over 1 + 1({1\over 4})})

= tan^{-1}({3\over 5})


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