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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