The area of the triangle formed by the coordi | vedclass

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(C) Given the curve $xy = a^2$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At the point $(

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$2a^2$

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(C) Given the curve $xy = a^2$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At the point $(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1}$.The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$.Multiplying by $x_1$,we get $x_1 y - x_1 y_1 = -y_1 x + x_1 y_1$,which simplifies to $x_1 y + y_1 x = 2x_1 y_1$.Since $x_1 y_1 = a^2$,the equation becomes $x_1 y + y_1 x = 2a^2$.Dividing by $2a^2$,we get $\frac{x}{2a^2/y_1} + \frac{y}{2a^2/x_1} = 1$.This is the intercept form of a line $\frac{x}{A} + \frac{y}{B} = 1$,where the x-intercept $A = \frac{2a^2}{y_1}$ and the y-intercept $B = \frac{2a^2}{x_1}$.The area of the triangle formed by the axes and the tangent is $\frac{1}{2} 	imes |A| 	imes |B| = \frac{1}{2} 	imes \frac{2a^2}{y_1} 	imes \frac{2a^2}{x_1} = \frac{2a^4}{x_1 y_1}$.Since $x_1 y_1 = a^2$,the area is $\frac{2a^4}{a^2} = 2a^2$ sq. units.

The area of the triangle formed by the coordinate axes and a tangent to the curve $xy = a^2$ at the point $(x_1, y_1)$ is . . . . . . sq. units (where $a, x_1$,and $y_1$ are non-zero).

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