An angle between the curves x^2 y = 1 and y(x | vedclass

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(D) Given curves are $C_1: x^2 y = 1$ and $C_2: y(x^2 + 1) = 2$. First,find the intersection points by substituting $y = 1/x^2$ into the second equati

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$\operatorname{Tan}^{-1} \frac{1}{3}$

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(D) Given curves are $C_1: x^2 y = 1$ and $C_2: y(x^2 + 1) = 2$. First,find the intersection points by substituting $y = 1/x^2$ into the second equation: $(1/x^2)(x^2 + 1) = 2 \implies 1 + 1/x^2 = 2 \implies 1/x^2 = 1 \implies x^2 = 1 \implies x = \pm 1$. For $x = 1$,$y = 1$. For $x = -1$,$y = 1$. Intersection points are $(1, 1)$ and $(-1, 1)$. For $C_1$,$y = x^{-2} \implies dy/dx = -2x^{-3} = -2/x^3$. At $(1, 1)$,$m_1 = -2$. For $C_2$,$y = 2/(x^2 + 1) \implies dy/dx = -2(2x)/(x^2 + 1)^2 = -4x/(x^2 + 1)^2$. At $(1, 1)$,$m_2 = -4(1)/(1+1)^2 = -4/4 = -1$. The angle $	heta$ between the curves is given by $	an 	heta = |(m_1 - m_2) / (1 + m_1 m_2)|$. $	an 	heta = |(-2 - (-1)) / (1 + (-2)(-1))| = |-1 / (1 + 2)| = |-1/3| = 1/3$. Thus,$	heta = \operatorname{Tan}^{-1}(1/3)$.

An angle between the curves $x^2 y = 1$ and $y(x^2 + 1) = 2$ is

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