If the tangent to the curve xy+ax+by=0 at (1, | vedclass

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(B) The given curve is $xy+ax+by=0$. Since the curve passes through $(1,1)$, we have $1(1)+a(1)+b(1)=0$, which implies $a+b=-1$ (Equation $1$). Differ

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$\frac{1}{2}$

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(B) The given curve is $xy+ax+by=0$. Since the curve passes through $(1,1)$, we have $1(1)+a(1)+b(1)=0$, which implies $a+b=-1$ (Equation $1$). Differentiating the equation of the curve with respect to $x$, we get $x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$. Rearranging for $\frac{dy}{dx}$, we get $\frac{dy}{dx}(x+b) = -(y+a)$, so $\frac{dy}{dx} = -\frac{y+a}{x+b}$. At the point $(1,1)$, the slope of the tangent is $m = \left(\frac{dy}{dx}ight)_{(1,1)} = -\frac{1+a}{1+b}$. Given that the tangent makes an angle $	an^{-1} 2$ with the $x$-axis, the slope $m = 	an(	an^{-1} 2) = 2$. Equating the two expressions for the slope: $-\frac{1+a}{1+b} = 2$, which simplifies to $-(1+a) = 2(1+b)$, or $a+2b = -3$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a+2b) - (a+b) = -3 - (-1)$, which gives $b = -2$. Substituting $b = -2$ into Equation $1$: $a - 2 = -1$, so $a = 1$. Finally, $\frac{a+b}{ab} = \frac{1+(-2)}{(1)(-2)} = \frac{-1}{-2} = \frac{1}{2}$.

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

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