If two parabolas y^2 = 4x and x^2 = 32y inter | vedclass

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(A) For the parabola $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At the point $(16,

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$tan^{-1}(3/5)$

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(A) For the parabola $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At the point $(16, 8)$,the slope $m_1 = \frac{2}{8} = \frac{1}{4}$.For the parabola $x^2 = 32y$,differentiating with respect to $x$ gives $2x = 32 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{16}$. At the point $(16, 8)$,the slope $m_2 = \frac{16}{16} = 1$.The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.Substituting the values,$	an 	heta = |\frac{1 - 1/4}{1 + (1)(1/4)}| = |\frac{3/4}{5/4}| = \frac{3}{5}$.Thus,$	heta = 	an^{-1}(3/5)$.

If two parabolas $y^2 = 4x$ and $x^2 = 32y$ intersect at the point $(16, 8)$ at an angle $\theta$,then what is the value of $\tan \theta$?

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