If the tangent to the curve xy + ax + by = 0 | vedclass

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(A) The point $(1, 1)$ lies on the curve $xy + ax + by = 0$,so $1(1) + a(1) + b(1) = 0$,which gives $a + b = -1$  .......$(1)$The slope of the tangent

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$1/2$

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(A) The point $(1, 1)$ lies on the curve $xy + ax + by = 0$,so $1(1) + a(1) + b(1) = 0$,which gives $a + b = -1$  .......$(1)$The slope of the tangent at $(1, 1)$ is given by $\frac{dy}{dx} = 	an(	an^{-1}(2)) = 2$.Differentiating the equation $xy + ax + by = 0$ with respect to $x$,we get:$x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$Substituting $x = 1$,$y = 1$,and $\frac{dy}{dx} = 2$ into the derivative equation:$1(2) + 1 + a + b(2) = 0$$3 + a + 2b = 0 \Rightarrow a + 2b = -3$  ........$(2)$Subtracting equation $(1)$ from equation $(2)$:$(a + 2b) - (a + b) = -3 - (-1)$$b = -2$Substituting $b = -2$ into equation $(1)$:$a - 2 = -1 \Rightarrow a = 1$Finally,calculating the required value:$\frac{a + b}{ab} = \frac{1 + (-2)}{1 	imes (-2)} = \frac{-1}{-2} = \frac{1}{2}$

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle $\tan^{-1}(2)$ with the $x$-axis,then $\left( \frac{a + b}{ab} \right)$ is equal to:

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