If frac{pi}{3} and theta are the eccentric an | vedclass

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(A) Given the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,we have $a^2=16$ and $b^2=12$.The eccentricity $e$ is given by $b^2=a^2(1-e^2)$,so $12=16(1-e^

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$-\sqrt{3}$

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(A) Given the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,we have $a^2=16$ and $b^2=12$.The eccentricity $e$ is given by $b^2=a^2(1-e^2)$,so $12=16(1-e^2)$,which gives $1-e^2=\frac{3}{4}$,so $e^2=\frac{1}{4}$ and $e=\frac{1}{2}$.For a focal chord with eccentric angles $\alpha$ and $\beta$,the condition is $	an(\frac{\alpha}{2}) 	an(\frac{\beta}{2}) = \frac{e-1}{e+1}$.Here $\alpha = \frac{\pi}{3}$,so $\frac{\alpha}{2} = \frac{\pi}{6}$.$	an(\frac{\pi}{6}) 	an(\frac{	heta}{2}) = \frac{1/2 - 1}{1/2 + 1} = \frac{-1/2}{3/2} = -\frac{1}{3}$.Since $	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} 	an(\frac{	heta}{2}) = -\frac{1}{3}$,so $	an(\frac{	heta}{2}) = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$.Thus,$\frac{	heta}{2} = -\frac{\pi}{6}$,which means $	heta = -\frac{\pi}{3}$.Therefore,$	an 	heta = 	an(-\frac{\pi}{3}) = -\sqrt{3}$.

If $\frac{\pi}{3}$ and $\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,then $\tan \theta=$

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