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

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(C) Step $1$: Since $(1, 1)$ lies on the curve $xy + ax + by = 0$, we have $(1)(1) + a(1) + b(1) = 0$, which implies $1 + a + b = 0$, or $a + b = -1$.

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$2$

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(C) Step $1$: Since $(1, 1)$ lies on the curve $xy + ax + by = 0$, we have $(1)(1) + a(1) + b(1) = 0$, which implies $1 + a + b = 0$, or $a + b = -1$.Step $2$: Differentiate the equation $xy + ax + by = 0$ with respect to $x$: $y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$.Step $3$: At $(1, 1)$, the slope $\frac{dy}{dx} = 	an(	an^{-1} 2) = 2$. Substituting these values: $1 + (1)(2) + a + b(2) = 0$.Step $4$: This simplifies to $1 + 2 + a + 2b = 0$, so $a + 2b = -3$.Step $5$: Solving the system $a + b = -1$ and $a + 2b = -3$, subtract the first from the second: $(a + 2b) - (a + b) = -3 - (-1) \implies b = -2$. Then $a = -1 - (-2) = 1$.Step $6$: Calculate $\frac{ab}{a + b} = \frac{(1)(-2)}{-1} = \frac{-2}{-1} = 2$.

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

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