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(C) Given the curve $xy = a^2$,we have $y = \frac{a^2}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2}$.At the

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$2a^2$

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(C) Given the curve $xy = a^2$,we have $y = \frac{a^2}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2}$.At the point $(x_1, y_1)$,the slope of the tangent is $m = -\frac{a^2}{x_1^2}$.The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{a^2}{x_1^2}(x - x_1)$.Multiplying by $x_1^2$,we get $x_1^2 y - x_1^2 y_1 = -a^2 x + a^2 x_1$.Since $x_1 y_1 = a^2$,this becomes $x_1^2 y + a^2 x = a^2 x_1 + x_1^2 y_1 = x_1(a^2 + x_1 y_1) = x_1(a^2 + a^2) = 2a^2 x_1$.Dividing by $a^2 x_1$,the equation is $\frac{x}{2x_1} + \frac{y}{2a^2/x_1} = 1$.This is the intercept form of a line $\frac{x}{a'} + \frac{y}{b'} = 1$,where the intercepts are $a' = 2x_1$ and $b' = \frac{2a^2}{x_1}$.The area of the triangle formed by the axes and the tangent is $\frac{1}{2} 	imes |a'| 	imes |b'| = \frac{1}{2} 	imes (2x_1) 	imes \left(\frac{2a^2}{x_1}ight) = 2a^2$.

The area of the triangle formed by the coordinate axes and a tangent to the curve $xy = a^2$ at the point $(x_1, y_1)$ on it is

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