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(B) The equation of the ellipse is $16 x^2 + 11 y^2 = 256$,which can be written as $\frac{x^2}{16} + \frac{y^2}{256/11} = 1$. The tangent at point $(x

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$\pm \frac{\pi}{6}$

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(B) The equation of the ellipse is $16 x^2 + 11 y^2 = 256$,which can be written as $\frac{x^2}{16} + \frac{y^2}{256/11} = 1$. The tangent at point $(x_1, y_1)$ is $\frac{x x_1}{16} + \frac{y y_1}{256/11} = 1$. Substituting $(x_1, y_1) = \left(4 \cos 2 	heta, \frac{16}{\sqrt{11}} \sin 2 	hetaight)$,we get $\frac{x(4 \cos 2 	heta)}{16} + \frac{y(16 \sin 2 	heta)}{\sqrt{11}(256/11)} = 1$,which simplifies to $x \cos 2 	heta + \frac{\sqrt{11}}{16} y \sin 2 	heta = 4$,or $4 x \cos 2 	heta + y \sqrt{11} \sin 2 	heta = 16$. The circle is $x^2 + y^2 - 2 x - 15 = 0$,with center $(1, 0)$ and radius $r = \sqrt{1^2 + 0^2 - (-15)} = 4$. The perpendicular distance from the center $(1, 0)$ to the tangent line must equal the radius $4$: $\frac{|4(1) \cos 2 	heta + 0 - 16|}{\sqrt{16 \cos^2 2 	heta + 11 \sin^2 2 	heta}} = 4$. Squaring both sides: $\frac{(4 \cos 2 	heta - 16)^2}{16 \cos^2 2 	heta + 11 \sin^2 2 	heta} = 16$. Dividing by $16$: $\frac{(\cos 2 	heta - 4)^2}{16 \cos^2 2 	heta + 11 \sin^2 2 	heta} = 1$. $\cos^2 2 	heta - 8 \cos 2 	heta + 16 = 16 \cos^2 2 	heta + 11(1 - \cos^2 2 	heta) = 5 \cos^2 2 	heta + 11$. Rearranging gives $4 \cos^2 2 	heta + 8 \cos 2 	heta - 5 = 0$. Solving for $\cos 2 	heta$ using the quadratic formula: $\cos 2 	heta = \frac{-8 \pm \sqrt{64 - 4(4)(-5)}}{8} = \frac{-8 \pm \sqrt{144}}{8} = \frac{-8 \pm 12}{8}$. Thus,$\cos 2 	heta = \frac{4}{8} = \frac{1}{2}$ or $\cos 2 	heta = -\frac{20}{8} = -2.5$ (rejected). So,$2 	heta = \pm \frac{\pi}{3}$,which gives $	heta = \pm \frac{\pi}{6}$.

If the tangent at the point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the ellipse $16 x^2+11 y^2=256$ touches the circle $x^2+y^2-2 x=15$,then $\theta=$

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