Let S be the focus of the hyperbola frac{x^2} | vedclass

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(C) For the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$,we have $a^2=3$ and $b^2=5$. The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{5}{3

Vedclass · Accepted Answer

$40$

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(C) For the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$,we have $a^2=3$ and $b^2=5$. The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{5}{3}} = \sqrt{\frac{8}{3}}$.The focus $S$ is $(ae, 0) = (\sqrt{3} \cdot \sqrt{\frac{8}{3}}, 0) = (\sqrt{8}, 0)$.The circle $C$ has center $A(\sqrt{6}, \sqrt{5})$ and passes through $S(\sqrt{8}, 0)$.The radius $r$ is the distance $AS = \sqrt{(\sqrt{8}-\sqrt{6})^2 + (0-\sqrt{5})^2} = \sqrt{8+6-2\sqrt{48}+5} = \sqrt{19-8\sqrt{3}}$.Since $SAB$ is a diameter,$B$ is the point such that $A$ is the midpoint of $SB$. Thus,$B = 2A - S = (2\sqrt{6}-\sqrt{8}, 2\sqrt{5})$.The triangle $OSB$ has vertices $O(0,0)$,$S(\sqrt{8}, 0)$,and $B(2\sqrt{6}-\sqrt{8}, 2\sqrt{5})$.The area of $	riangle OSB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes OS 	imes y_B = \frac{1}{2} 	imes \sqrt{8} 	imes 2\sqrt{5} = \sqrt{40}$.The square of the area is $(\sqrt{40})^2 = 40$.

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