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

p = 2, q = -7  p = -2, q = 7  p = -2, q = -7  p = 2, q = 7

p = 2, q = -7

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