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

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(C) Given the curve equation is $y^2 = px^3 + q \dots (i)$.Since the point $(2, 3)$ lies on the curve,we substitute $x=2$ and $y=3$ into $(i)$:$3^2 =

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$2, -7$

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(C) Given the curve equation is $y^2 = px^3 + q \dots (i)$.Since the point $(2, 3)$ lies on the curve,we substitute $x=2$ and $y=3$ into $(i)$:$3^2 = p(2)^3 + q \Rightarrow 9 = 8p + q \dots (ii)$.Differentiating both sides of $(i)$ with respect to $x$:$2y \frac{dy}{dx} = 3px^2 \Rightarrow \frac{dy}{dx} = \frac{3px^2}{2y}$.The slope of the tangent at $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$.The given tangent line is $y = 4x - 5$,which has a slope of $4$.Equating the slopes: $2p = 4 \Rightarrow p = 2$.Substituting $p = 2$ into $(ii)$:$9 = 8(2) + q \Rightarrow 9 = 16 + q \Rightarrow q = -7$.Thus,the values are $p = 2$ and $q = -7$.

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

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