# 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

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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