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Question

(B) Given the hyperbola $xy = a^2$,which can be written as $y = \frac{a^2}{x}$.Let the point of tangency be $(x_0, y_0)$. The derivative is $\frac{dy}

Vedclass · Accepted Answer

$2a^2$

Vedclass · Answer

(B) Given the hyperbola $xy = a^2$,which can be written as $y = \frac{a^2}{x}$.Let the point of tangency be $(x_0, y_0)$. The derivative is $\frac{dy}{dx} = -\frac{a^2}{x^2}$.At any point $(x_0, y_0)$ on the curve,the slope of the tangent is $m = -\frac{a^2}{x_0^2}$.Since $y_0 = \frac{a^2}{x_0}$,the equation of the tangent is $y - y_0 = -\frac{a^2}{x_0^2}(x - x_0)$.Substituting $y_0 = \frac{a^2}{x_0}$,we get $y - \frac{a^2}{x_0} = -\frac{a^2}{x_0^2}x + \frac{a^2}{x_0}$.This simplifies to $y = -\frac{a^2}{x_0^2}x + \frac{2a^2}{x_0}$,or $\frac{x}{x_0} + \frac{y}{y_0} = 2$.The $x$-intercept is $2x_0$ and the $y$-intercept is $2y_0$.The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{1}{2} 	imes |2x_0| 	imes |2y_0| = 2|x_0 y_0|$.Since $x_0 y_0 = a^2$,the area is $2|a^2| = 2a^2$.

The area of the triangle formed by the tangent to the hyperbola $xy = a^2$ and the coordinate axes is:

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