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(C) The given parabolas are $y^2=32x$ $(i)$ and $2x^2=27y$ (ii).To find the intersection points,substitute $x = \frac{y^2}{32}$ into (ii):$2(\frac{y^2

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$3\sqrt{2}$

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(C) The given parabolas are $y^2=32x$ $(i)$ and $2x^2=27y$ (ii).To find the intersection points,substitute $x = \frac{y^2}{32}$ into (ii):$2(\frac{y^2}{32})^2 = 27y \implies 2 \cdot \frac{y^4}{1024} = 27y \implies \frac{y^4}{512} = 27y \implies y(y^3 - 512 \cdot 27) = 0$.Thus,$y=0$ or $y^3 = (8^3)(3^3) = 24^3$,so $y=24$.For $y=24$,$x = \frac{24^2}{32} = \frac{576}{32} = 18$. So $P = (18, 24)$.For $y^2=32x$,differentiating gives $2y \frac{dy}{dx} = 32 \implies \frac{dy}{dx} = \frac{16}{y}$. At $P(18, 24)$,$m_1 = \frac{16}{24} = \frac{2}{3}$.For $2x^2=27y$,differentiating gives $4x = 27 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{4x}{27}$. At $P(18, 24)$,$m_2 = \frac{4(18)}{27} = \frac{72}{27} = \frac{8}{3}$.The tangent of the angle $	heta$ is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{8/3 - 2/3}{1 + (8/3)(2/3)}| = |\frac{6/3}{1 + 16/9}| = |\frac{2}{25/9}| = \frac{18}{25}$.Therefore,$5\sqrt{	an 	heta} = 5\sqrt{\frac{18}{25}} = 5 \cdot \frac{3\sqrt{2}}{5} = 3\sqrt{2}$.

If $P$ and the origin are the points of intersection of the parabolas $y^2=32x$ and $2x^2=27y$,and if $\theta$ is the acute angle between these curves at $P$,then $5\sqrt{\tan \theta} =$

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