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
Let \(\theta\) be the angle, then angle between angle between two curves
\(tan \theta\) = \(m_1 – m_2\over 1 + m_1 m_2\) = \(3\over 11\)
\(\theta\) = \(tan^{-1} (3/11)\)
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