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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The focus $S$ is $(ae, 0)$.The line is $bx-ay=0$,which can be writte

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$ab$

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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The focus $S$ is $(ae, 0)$.The line is $bx-ay=0$,which can be written as $bx-ay+0=0$.The length of the perpendicular $SP$ from $S(ae, 0)$ to the line $bx-ay=0$ is:$SP = \left| \frac{b(ae) - a(0)}{\sqrt{b^{2}+(-a)^{2}}} ight| = \frac{abe}{\sqrt{b^{2}+a^{2}}}$.Since $b^{2} = a^{2}(e^{2}-1)$,we have $a^{2}+b^{2} = a^{2}e^{2}$,so $\sqrt{a^{2}+b^{2}} = ae$.Thus,$SP = \frac{abe}{ae} = b$.The distance $CS$ is $ae$. In the right-angled triangle $\Delta SPC$,$CP^{2} = CS^{2} - SP^{2}$.$CP^{2} = (ae)^{2} - b^{2} = a^{2}e^{2} - b^{2} = a^{2}(1 + \frac{b^{2}}{a^{2}}) - b^{2} = a^{2} + b^{2} - b^{2} = a^{2}$.Therefore,$CP = a$.The area of the rectangle with sides $SP$ and $CP$ is $SP 	imes CP = b 	imes a = ab$.

Let $P$ be the foot of the perpendicular from the focus $S$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ on the line $bx-ay=0$ and let $C$ be the centre of the hyperbola. Then,the area of the rectangle whose sides are equal to $SP$ and $CP$ is

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