If the line 6x - y - 4 = 0 touches the curve | vedclass

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(C) The slope of the line $6x - y - 4 = 0$ is $6$. Since this line is tangent to the curve $y^{2} = ax^{3} + b$ at the point $(1, 2)$,the derivative a

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$4$

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(C) The slope of the line $6x - y - 4 = 0$ is $6$. Since this line is tangent to the curve $y^{2} = ax^{3} + b$ at the point $(1, 2)$,the derivative at this point must equal the slope of the line.Differentiating the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^{2} \Rightarrow \frac{dy}{dx} = \frac{3ax^{2}}{2y}$.At the point $(1, 2)$,the slope is $\left(\frac{dy}{dx}\right)_{(1, 2)} = \frac{3a(1)^{2}}{2(2)} = \frac{3a}{4}$.Equating the slope to $6$: $\frac{3a}{4} = 6 \Rightarrow 3a = 24 \Rightarrow a = 8$.Since the point $(1, 2)$ lies on the curve,it must satisfy the equation:$(2)^{2} = a(1)^{3} + b \Rightarrow 4 = 8(1) + b \Rightarrow b = 4 - 8 = -4$.Therefore,$a + b = 8 + (-4) = 4$.

If the line $6x - y - 4 = 0$ touches the curve $y^{2} = ax^{3} + b$ at the point $(1, 2)$,then $a + b =$

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