If the slope of a tangent to the curve xy + a | vedclass

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(A) Given the curve is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) = 0 \impl

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$3$

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(A) Given the curve is $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) = 0 \implies a + b = -1$ ... $(i)$ Now,differentiate the equation $xy + ax + by = 0$ with respect to $x$: $x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$ $(x + b) \frac{dy}{dx} = -(y + a)$ $\frac{dy}{dx} = -\frac{y + a}{x + b}$ Given that the slope at $(1, 1)$ is $2$: $2 = -\frac{1 + a}{1 + b}$ $2(1 + b) = -(1 + a)$ $2 + 2b = -1 - a$ $a + 2b = -3$ ... $(ii)$ Subtracting equation $(i)$ from equation $(ii)$: $(a + 2b) - (a + b) = -3 - (-1)$ $b = -2$ Substituting $b = -2$ into equation $(i)$: $a - 2 = -1 \implies a = 1$ Therefore,$a - b = 1 - (-2) = 3$.

If the slope of a tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ on it is $2$,then $a - b$ is

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