If the line y = 4x - 5 is tangent to the curv | vedclass

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

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$28$

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(D) Step $1$: Since the point $(2, 3)$ lies on the curve $y^2 = ax^3 + b$, we have $3^2 = a(2)^3 + b$, which simplifies to $9 = 8a + b$ (Equation $1$).Step $2$: Differentiate the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$, so $\frac{dy}{dx} = \frac{3ax^2}{2y}$.Step $3$: The slope of the tangent at $(2, 3)$ is $\frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$. Given the line $y = 4x - 5$ has a slope of $4$, we equate $2a = 4$, giving $a = 2$.Step $4$: Substitute $a = 2$ into Equation $1$: $9 = 8(2) + b \implies 9 = 16 + b \implies b = -7$.Step $5$: Calculate $7a - 2b = 7(2) - 2(-7) = 14 + 14 = 28$.

If the line $y = 4x - 5$ is tangent to the curve $y^2 = ax^3 + b$ at the point $(2, 3)$, then the value of $7a - 2b$ is...

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