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(C) Let the point $P$ be $(at, \frac{a}{t})$.The equation of the curve is $xy = a^2$.Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} =

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$2a^2$ sq. units

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(C) Let the point $P$ be $(at, \frac{a}{t})$.The equation of the curve is $xy = a^2$.Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At point $P(at, \frac{a}{t})$,the slope of the tangent is $m = -\frac{a/t}{at} = -\frac{1}{t^2}$.The equation of the tangent at $P$ is $y - \frac{a}{t} = -\frac{1}{t^2}(x - at)$.$t^2y - at = -x + at$,which simplifies to $x + t^2y = 2at$.To find the $x$-intercept,set $y = 0$: $x = 2at$. So,$A = (2at, 0)$.To find the $y$-intercept,set $x = 0$: $t^2y = 2at$,so $y = \frac{2a}{t}$. So,$B = (0, \frac{2a}{t})$.The area of the right-angled triangle $	riangle ABO$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2at) 	imes (\frac{2a}{t}) = 2a^2$ sq. units.

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

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