The angle of intersection between the curve \(x^2\) = 32y and \(y^2\) = 4x at point (16, 8) is

Solution :

\(x^2\) = 32y \(\implies\) \(dy\over dx\) = \(x\over 16\) \(\implies\) \(y^2\) = 4x \(\implies\) \(dy\over dx\) = \(2\over y\)

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

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

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