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

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(A) Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we have:$3^2 = p(2)^3 + q$$9 = 8p + q$  ...$(i)$Now,differentiate the curve equation $y^2=p

Vedclass · Accepted Answer

$p=2, q=-7$

Vedclass · Answer

(A) Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we have:$3^2 = p(2)^3 + q$$9 = 8p + q$  ...$(i)$Now,differentiate the curve equation $y^2=px^3+q$ with respect to $x$:$2y \frac{dy}{dx} = 3px^2$$\frac{dy}{dx} = \frac{3px^2}{2y}$The slope of the tangent $y=4x-5$ is $4$. Therefore,the derivative at $(2,3)$ must be equal to $4$:$\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3p(2)^2}{2(3)} = 4$$\frac{12p}{6} = 4$$2p = 4 \Rightarrow p = 2$  ...$(ii)$Substitute $p=2$ into equation $(i)$:$8(2) + q = 9$$16 + q = 9$$q = 9 - 16 = -7$Thus,$p=2$ and $q=-7$.

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

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