If y = 4x - 5 is tangent to the curve y^2 = p | vedclass

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

Vedclass · Accepted Answer

$p = 2, q = -7$

Vedclass · Answer

(A) Given the curve $y^2 = px^3 + q$  $(i)$Differentiating with respect to $x$,we get $2y \cdot \frac{dy}{dx} = 3px^2$.Thus,$\frac{dy}{dx} = \frac{3px^2}{2y}$.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,$2p = 4$,which gives $p = 2$.Since the point $(2, 3)$ lies on the curve,substitute $x = 2, y = 3$ and $p = 2$ into equation $(i)$:$3^2 = 2(2)^3 + q$$9 = 2(8) + q$$9 = 16 + q$$q = 9 - 16 = -7$.Therefore,$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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