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

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Question

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

Vedclass · Accepted Answer

$a = 1, b = 2$

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

If the slope of the tangent to the curve $xy + ax - by = 0$ at the point $(1, 1)$ is $2$,then what are the values of $a$ and $b$ respectively?

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