The area of the triangle formed by any tangen | vedclass

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(B) Let the point of tangency be $P(x_1, y_1)$ on the curve $xy = 100$. Since $P$ lies on the curve,$y_1 = \frac{100}{x_1}$. Differentiating $xy = 100

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) Let the point of tangency be $P(x_1, y_1)$ on the curve $xy = 100$. Since $P$ lies on the curve,$y_1 = \frac{100}{x_1}$. Differentiating $xy = 100$ with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$. At point $P(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1} = -\frac{100/x_1}{x_1} = -\frac{100}{x_1^2}$. The equation of the tangent at $P(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$. $y - y_1 = -\frac{y_1}{x_1}x + y_1 \implies \frac{y_1}{x_1}x + y = 2y_1$. Dividing by $2y_1$,we get $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$. The $x$-intercept is $A(2x_1, 0)$ and the $y$-intercept is $B(0, 2y_1)$. The area of $\Delta OAB = \frac{1}{2} 	imes |2x_1| 	imes |2y_1| = 2|x_1 y_1|$. Since $x_1 y_1 = 100$,the area is $2 	imes 100 = 200$ square units.

The area of the triangle formed by any tangent to the curve $xy = 100$ with the coordinate axes is:

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