The angle between the curves y^2 = 4x and x^2 | vedclass

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(A) For the curve $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$.At the point $(1, 2)$,t

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$	an^{-1}(3)$

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(A) For the curve $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$.At the point $(1, 2)$,the slope $m_1 = \frac{2}{2} = 1$.For the curve $x^2 + y^2 = 5$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.At the point $(1, 2)$,the slope $m_2 = -\frac{1}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} ight| = \left| \frac{3/2}{1/2} ight| = 3$.Therefore,$	heta = 	an^{-1}(3)$.

The angle between the curves $y^2 = 4x$ and $x^2 + y^2 = 5$ at the point $(1, 2)$ is:

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