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

$\frac{dy}{dx}(2,3)=\frac{3px^2}{2y}=\frac{12p}{6}=2p$ So, 2p = 4  ⇒ p = 2  $∴\frac{dy}{dx}=4$

Also, (2, 3) lies on $y^2=px^3+q⇒9=2(2)^3+q$ 

⇒ q = -7