Normals are drawn from the point P(8,0) to th | vedclass

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(B) Given parabola is $y^2=12x$,so $4a=12 \Rightarrow a=3$. Equation of normal with slope $m$ is $y=mx-2am-am^3$,which becomes $y=mx-6m-3m^3$. Since t

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$2 \sqrt{6}$

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(B) Given parabola is $y^2=12x$,so $4a=12 \Rightarrow a=3$. Equation of normal with slope $m$ is $y=mx-2am-am^3$,which becomes $y=mx-6m-3m^3$. Since the normal passes through $P(8,0)$,we have $0=8m-6m-3m^3$. $2m-3m^3=0 \Rightarrow m(2-3m^2)=0$. The slopes are $m_1=0$,$m_2=\sqrt{\frac{2}{3}}$,and $m_3=-\sqrt{\frac{2}{3}}$. The non-horizontal normals have slopes $m_2=\sqrt{\frac{2}{3}}$ and $m_3=-\sqrt{\frac{2}{3}}$. The angle $	heta$ between these two normals is given by $	an 	heta = \left| \frac{m_2-m_3}{1+m_2m_3} ight|$. $	an 	heta = \left| \frac{\sqrt{\frac{2}{3}} - (-\sqrt{\frac{2}{3}})}{1 + (\sqrt{\frac{2}{3}})(-\sqrt{\frac{2}{3}})} ight| = \left| \frac{2\sqrt{\frac{2}{3}}}{1 - \frac{2}{3}} ight| = \frac{2\sqrt{\frac{2}{3}}}{\frac{1}{3}} = 6 \sqrt{\frac{2}{3}} = 6 \frac{\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$.

Normals are drawn from the point $P(8,0)$ to the parabola $y^2=12x$. If $\theta$ is the acute angle between two non-horizontal normals among them,then $\tan \theta=$

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