Given curve is y2=px3+q....(i) On differentiating w.r.t. x, we get 2y
dy
dx
=3px2⇒
dy
dx
=
3px2
2y
At (2,3),(
dy
dx
)(2,3)=
12p
6
=2p since, line y=4x−5 is a tangent to the given curve. ∴ Slope =4=2p ⇒p=2 Now, put the value of p,x and y in Eq. (i), we get (3)2=(2)(2)3+q ⇒9=16+q ⇒q=−7 Hence, p+q=2−7=−5