If y=4x-5 is a tangent to the curve y^2=px^3+ | vedclass

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(C) $y^2=px^3+q$ ... $(i)$ Differentiating both sides with respect to $x$,we get $2y \cdot \frac{dy}{dx} = 3px^2$ $\frac{dy}{dx} = \frac{3px^2}{2y}$ A

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$9$

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(C) $y^2=px^3+q$ ... $(i)$ Differentiating both sides with respect to $x$,we get $2y \cdot \frac{dy}{dx} = 3px^2$ $\frac{dy}{dx} = \frac{3px^2}{2y}$ At the point $(2,3)$,the slope of the tangent is: $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$ The slope of the line $y=4x-5$ is $4$. Since the line is tangent to the curve,their slopes must be equal: $2p = 4 \Rightarrow p = 2$ Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we substitute the values: $3^2 = p(2)^3 + q$ $9 = 8p + q$ Substituting $p=2$: $9 = 8(2) + q$ $9 = 16 + q \Rightarrow q = -7$ Therefore,$p-q = 2 - (-7) = 2 + 7 = 9$.

If $y=4x-5$ is a tangent to the curve $y^2=px^3+q$ at $(2,3)$,then $p-q$ is

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