If y=2x is a tangent to the curve y^2=ax^3+b | vedclass

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Question

(C) Given the curve $y^2 = ax^3 + b$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 3ax^2$,which implies $\frac{dy}{dx} = \frac{3ax^2

Vedclass · Accepted Answer

$(\frac{8}{3}, \frac{4}{3})$

Vedclass · Answer

(C) Given the curve $y^2 = ax^3 + b$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 3ax^2$,which implies $\frac{dy}{dx} = \frac{3ax^2}{2y}$. The slope of the tangent at $(1,2)$ is $\left. \frac{dy}{dx} \right|_{(1,2)} = \frac{3a(1)^2}{2(2)} = \frac{3a}{4}$. The given tangent is $y = 2x$,which has a slope of $2$. Equating the slopes,$\frac{3a}{4} = 2$,so $a = \frac{8}{3}$. Since the curve passes through $(1,2)$,we substitute these values into the curve equation: $2^2 = \frac{8}{3}(1)^3 + b$. $4 = \frac{8}{3} + b$,which gives $b = 4 - \frac{8}{3} = \frac{4}{3}$. Therefore,$(a, b) = (\frac{8}{3}, \frac{4}{3})$.

If $y=2x$ is a tangent to the curve $y^2=ax^3+b$ at $(1,2)$,then $(a, b)=$

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