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(B) The given equation of the hyperbola is $\frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1$. Differentiating with respect to $x$,we get $2(x-1) - (y-2) \frac{d

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$4$

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(B) The given equation of the hyperbola is $\frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1$. Differentiating with respect to $x$,we get $2(x-1) - (y-2) \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{2(x-1)}{y-2}$. The equation of the tangent is given as $x=2$,which is a vertical line. For a vertical tangent,the slope $\frac{dy}{dx}$ must be undefined,meaning the denominator $y-2 = 0$,so $y=2$. Since the point $(h, k)$ lies on the tangent $x=2$,we have $h=2$. Since the point $(h, k)$ lies on the hyperbola,substituting $y=k=2$ into the hyperbola equation gives $\frac{(x-1)^2}{1} - 0 = 1$,so $(x-1)^2 = 1$,which means $x-1 = \pm 1$. Thus $x=2$ or $x=0$. Since $h=2$,we have $k=2$. Therefore,$h+k = 2+2 = 4$.

If the equation of the tangent drawn at $(h, k)$ to the hyperbola $\frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1$ is $x=2$,then $h+k=$

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