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

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(B) Given the curve equation is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x=1$ and $y=1$ into the equation: $1(1)

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$2$

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(B) Given the curve equation is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x=1$ and $y=1$ into the equation: $1(1) + a(1) + b(1) = 0 \implies 1 + a + b = 0 \implies a + b = -1$. Now,differentiate the equation with respect to $x$: $y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$. Rearranging for $\frac{dy}{dx}$: $\frac{dy}{dx}(x + b) = -(y + a) \implies \frac{dy}{dx} = -\frac{y + a}{x + b}$. At the point $(1, 1)$,the slope of the tangent is $m = 	an(	an^{-1}(2)) = 2$. Substituting $(1, 1)$ into the derivative: $2 = -\frac{1 + a}{1 + b} \implies 2(1 + b) = -(1 + a) \implies 2 + 2b = -1 - a \implies a + 2b = -3$. We have a system of two linear equations: $1) a + b = -1$ $2) a + 2b = -3$ Subtracting $(1)$ from $(2)$: $(a + 2b) - (a + b) = -3 - (-1) \implies b = -2$. Substituting $b = -2$ into $(1)$: $a - 2 = -1 \implies a = 1$. Finally,calculate $\frac{ab}{a+b}$: $\frac{(1)(-2)}{1 + (-2)} = \frac{-2}{-1} = 2$.

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

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