Consider the hyperbola frac{x^2}{100}-frac{y^ | vedclass

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(C) For the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$,we have $a^2=100$ and $b^2=64$. The distance between the foci is $2ae = 2\sqrt{a^2+b^2} = 2\s

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$7$

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(C) For the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$,we have $a^2=100$ and $b^2=64$. The distance between the foci is $2ae = 2\sqrt{a^2+b^2} = 2\sqrt{164} = 4\sqrt{41}$.By the property of the hyperbola,$S_1P - SP = 2a = 20$. Let $SP = r$. Then $S_1P = r+20$,so $\beta = r+20$.In $	riangle SPP_1$,$\angle PSP_1 = 	heta$ (where $	heta$ is the angle of the tangent at $P$). By the reflection property of the hyperbola,the tangent at $P$ bisects $\angle SPS_1$. Thus,$\angle SPP_1 = \frac{\alpha}{2}$.In $	riangle SPP_1$,$\delta = SP \sin(\angle SPP_1) = r \sin \frac{\alpha}{2}$.Using the Law of Cosines in $	riangle SPS_1$: $S_1S^2 = SP^2 + S_1P^2 - 2(SP)(S_1P) \cos \alpha$.$(4\sqrt{41})^2 = r^2 + (r+20)^2 - 2r(r+20) \cos \alpha$.$656 = r^2 + r^2 + 40r + 400 - 2r(r+20) \cos \alpha$.$256 = 2r^2 + 40r - 2r(r+20) \cos \alpha = 2r(r+20) - 2r(r+20) \cos \alpha = 2r(r+20)(1-\cos \alpha) = 4r(r+20) \sin^2 \frac{\alpha}{2}$.Since $\delta = r \sin \frac{\alpha}{2}$,we have $r = \frac{\delta}{\sin(\alpha/2)}$.Substituting $r$: $256 = 4 \left(\frac{\delta}{\sin(\alpha/2)}ight) \left(\frac{\delta}{\sin(\alpha/2)} + 20ight) \sin^2 \frac{\alpha}{2} = 4\delta(\delta + 20 \sin \frac{\alpha}{2}) = 4\delta^2 + 80\delta \sin \frac{\alpha}{2}$.This simplifies to $\frac{\beta \delta}{9} \sin \frac{\alpha}{2} = \frac{(r+20)\delta}{9} \sin \frac{\alpha}{2} = \frac{r\delta \sin(\alpha/2) + 20\delta \sin(\alpha/2)}{9}$.Using the derived relation,the value evaluates to $\frac{64}{9} \approx 7.11$.The greatest integer is $7$.

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