If $\mathrm{y}=4\mathrm{x}-5$ is a tangent to the curve ${\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}$ 

at (2,3) then $|\mathrm{p}+\mathrm{q}|=$

Given (2,3) lies on ${\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}$ we get 9 = 8p + q ……..(1)
Again ${\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}\Rightarrow 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{px}}^2\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3{\mathrm{px}}^2}{2\mathrm{y}}$
$\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2,3)}=\frac{12\mathrm{p}}{6}=2\mathrm{p}$

 since $\mathrm{y}=4\mathrm{x}-5$ is tangent to ${\mathrm{y}}^2={\mathrm{px}}^3+\mathrm{q}$ at (2,3) therefore 2p = 4 $[\therefore$Slope of line $\mathrm{y}=4\mathrm{x}-5\text{ is }4$]

 $\Rightarrow \mathrm{p}=2$ Putting p=2 in equation (1), we get $\mathrm{q}=-7$Hence p = 2, q = -7.