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(B) Let the line perpendicular to the $X$-axis be $x = h = 3 \cos \alpha$. The point $A$ on the circle $x^2+y^2=9$ is $(3 \cos \alpha, 3 \sin \alpha)$

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$\frac{1}{2 \sqrt{6}}$

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(B) Let the line perpendicular to the $X$-axis be $x = h = 3 \cos \alpha$. The point $A$ on the circle $x^2+y^2=9$ is $(3 \cos \alpha, 3 \sin \alpha)$. The tangent at $A$ is $x(3 \cos \alpha) + y(3 \sin \alpha) = 9$,which simplifies to $x \cos \alpha + y \sin \alpha = 3$. The slope $m_1 = -\cot \alpha$.The point $B$ on the ellipse $4x^2+9y^2=36$ is $(3 \cos \alpha, 2 \sin \alpha)$. The tangent at $B$ is $4x(3 \cos \alpha) + 9y(2 \sin \alpha) = 36$,which simplifies to $2x \cos \alpha + 3y \sin \alpha = 6$. The slope $m_2 = -\frac{2}{3} \cot \alpha$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{-\cot \alpha + \frac{2}{3} \cot \alpha}{1 + \frac{2}{3} \cot^2 \alpha} ight| = \frac{\frac{1}{3} \cot \alpha}{1 + \frac{2}{3} \cot^2 \alpha} = \frac{\cot \alpha}{3 + 2 \cot^2 \alpha}$.To maximize $f(\cot \alpha) = \frac{\cot \alpha}{3 + 2 \cot^2 \alpha}$,let $u = \cot \alpha$. Then $f(u) = \frac{u}{3 + 2u^2}$.Setting $f'(u) = 0$,we get $\frac{(3 + 2u^2)(1) - u(4u)}{(3 + 2u^2)^2} = 0$,which implies $3 - 2u^2 = 0$,so $u^2 = \frac{3}{2}$ or $u = \sqrt{\frac{3}{2}}$.Substituting $u = \sqrt{\frac{3}{2}}$ into $f(u)$,we get $	an 	heta = \frac{\sqrt{3/2}}{3 + 2(3/2)} = \frac{\sqrt{3/2}}{6} = \frac{\sqrt{3}}{6 \sqrt{2}} = \frac{\sqrt{3}}{6 \sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{12} = \frac{1}{2 \sqrt{6}}$.

$A$ line perpendicular to the $X$-axis cuts the circle $x^2+y^2=9$ at $A$ and the ellipse $4x^2+9y^2=36$ at $B$ such that $A$ and $B$ lie in the same quadrant. If $\theta$ is the greatest acute angle between the tangents drawn to the curves at $A$ and $B$,then $\tan \theta=$

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