If the line y=4x-5 touches the curve y^2=ax^3 | vedclass

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(A) Given curve: $y^2 = ax^3 + b$. Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which simplifies to $9 = 8a + b$ (Equation $1$

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(A) Given curve: $y^2 = ax^3 + b$. Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which simplifies to $9 = 8a + b$ (Equation $1$). Differentiating the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$,so $\frac{dy}{dx} = \frac{3ax^2}{2y}$. At the point $(2,3)$,the slope of the tangent is $\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$. The given line is $y = 4x - 5$,which has a slope of $4$. Equating the slopes: $2a = 4$,which gives $a = 2$. Substituting $a = 2$ into Equation $1$: $9 = 8(2) + b \Rightarrow 9 = 16 + b \Rightarrow b = -7$. Now,calculate $7a + 2b$: $7(2) + 2(-7) = 14 - 14 = 0$.

If the line $y=4x-5$ touches the curve $y^2=ax^3+b$ at the point $(2,3)$,then $7a+2b=$

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