The angle between the curves xy=6 and x^2y=12 | vedclass

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(A) Given curves are $xy=6$ $(1)$ and $x^2y=12$ $(2)$. Dividing $(2)$ by $(1)$,we get $\frac{x^2y}{xy} = \frac{12}{6}$,which implies $x=2$. Substituti

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$	an^{-1} \frac{3}{11}$

Vedclass · Answer

(A) Given curves are $xy=6$ $(1)$ and $x^2y=12$ $(2)$. Dividing $(2)$ by $(1)$,we get $\frac{x^2y}{xy} = \frac{12}{6}$,which implies $x=2$. Substituting $x=2$ in $(1)$,we get $2y=6$,so $y=3$. The point of intersection is $(2, 3)$. For curve $(1)$,$y = \frac{6}{x}$,so $\frac{dy}{dx} = -\frac{6}{x^2}$. At $(2, 3)$,$m_1 = -\frac{6}{4} = -\frac{3}{2}$. For curve $(2)$,$y = \frac{12}{x^2}$,so $\frac{dy}{dx} = -\frac{24}{x^3}$. At $(2, 3)$,$m_2 = -\frac{24}{8} = -3$. The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. $	an 	heta = |\frac{-1.5 - (-3)}{1 + (-1.5)(-3)}| = |\frac{1.5}{1 + 4.5}| = |\frac{1.5}{5.5}| = \frac{15}{55} = \frac{3}{11}$. Therefore,$	heta = 	an^{-1} \frac{3}{11}$.

The angle between the curves $xy=6$ and $x^2y=12$ is

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