The tangent to the curve xy = 25 at any point | vedclass

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(A) Given curve is $xy = 25$. Let the point of tangency be $P(x_1, y_1)$. Since $P$ lies on the curve,$x_1 y_1 = 25$.Differentiating $xy = 25$ with re

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$50$ sq units

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(A) Given curve is $xy = 25$. Let the point of tangency be $P(x_1, y_1)$. Since $P$ lies on the curve,$x_1 y_1 = 25$.Differentiating $xy = 25$ with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.At $(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1}$.The equation of the tangent 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 $x_1 y + y_1 x = 2 x_1 y_1$.Dividing by $2 x_1 y_1$,we get $\frac{x}{2 x_1} + \frac{y}{2 y_1} = 1$.The tangent cuts the $x$-axis at $A(2 x_1, 0)$ and the $y$-axis at $B(0, 2 y_1)$.The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2 x_1) 	imes (2 y_1) = 2 x_1 y_1$.Since $x_1 y_1 = 25$,the area is $2(25) = 50$ sq units.

The tangent to the curve $xy = 25$ at any point on it cuts the coordinate axes at $A$ and $B$. Then the area of the $\triangle OAB$ is

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