For the hyperbola x^2-y^2-4x+2y+c=0,if the fo | vedclass

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(B) The given equation of the hyperbola is $x^2-y^2-4x+2y+c=0$.Completing the square,we get $(x-2)^2 - (y-1)^2 = 3-c$.Let $a^2 = 3-c$. The equation be

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(B) The given equation of the hyperbola is $x^2-y^2-4x+2y+c=0$.Completing the square,we get $(x-2)^2 - (y-1)^2 = 3-c$.Let $a^2 = 3-c$. The equation becomes $\frac{(x-2)^2}{a^2} - \frac{(y-1)^2}{a^2} = 1$.For this rectangular hyperbola,the eccentricity $e = \sqrt{2}$.The focus is given by $x = 2 \pm ae = 2 \pm \sqrt{a^2} \cdot \sqrt{2} = 2 \pm \sqrt{2a^2}$.Given the focus $S(2+2\sqrt{2}, k)$,we have $\sqrt{2a^2} = 2\sqrt{2} \implies 2a^2 = 8 \implies a^2 = 4$.Since $a^2 = 3-c$,we have $4 = 3-c$,which gives $c = -1$.The directrix is $x = 2 \pm \frac{a}{e} = 2 \pm \frac{2}{\sqrt{2}} = 2 \pm \sqrt{2}$.The directrix adjacent to the focus $x = 2+2\sqrt{2}$ is $x = 2+\sqrt{2}$,which matches the given condition.Thus,$c = -1$.

For the hyperbola $x^2-y^2-4x+2y+c=0$,if the focus is $S(2+2\sqrt{2}, k)$ and the directrix that is adjacent to $S$ is $x=2+\sqrt{2}$,then $c=$

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