Find the angle between the curves xy = 6 and \(x^2 y\) =12.

Solution :

The equation of the two curves are

xy = 6 …….(i)

and, \(x^2 y\) = 12 …………(ii)

from (i) , we obtain y = \(6\over x\). Putting this value of y in (ii), we obtain

\(x^2\) \((6\over x)\) = 12 \(\implies\) 6x = 12

\(\implies\) x = 2

Putting x = 2 in (i) or (ii), we get y = 3.

Thus, the two curves intersect at P(2, 3).

Differentiating (i) with respect to x, we get

x\(dy\over dx\) + y = 0 \(\implies\) \(dy\over dx\) = \(-y\over x\)

\(\implies\) \(m_1\) = \(({dy\over dx})_{(2, 3)}\) = \(-3\over 2\)

Differentiating (ii) with respect to x, we get

\(x^2\) \(dy\over dx\) + 2xy = 0 \(\implies\) \(dy\over dx\) = \(-2y\over x\)

\(\implies\) \(m_2\) = \(({dy\over dx})_{(2, 3)}\) = -3

\(tan \theta\) = \(m_1 – m_2\over 1 + m_1 m_2\) = \(3\over 11\)

\(\theta\) = \(tan^{-1} (3/11)\)