If $y=4x-5$ is a tangent to the curve $y^2=p{x}^3+q\text{ at }(2,3)$ then $pq=$

Point (2, 3) lies on $y^2=p{x}^3+q.$ . Therefore, 

$9=8p+q$

Now, 

  $\begin{array}{l}{y}^2=p{x}^3+q\\ \Rightarrow 2y\frac{dy}{dx}=3p{x}^2\\ \Rightarrow \frac{dy}{dx}=\frac{3p{x}^2}{2y}\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,3)}=\frac{12p}{6}=2p.\end{array}$

Since , $y=4x-5$ is $y^2=p{x}^3+q\text{ at }(2,3)$ 

$\therefore {\left(\frac{dy}{dx}\right)}_{(2,3)}=$ (Slope of the line $y=4x-5)$

$\Rightarrow 2p=4\Rightarrow p=2$

Putting $p = 2 in \left(i\right)$ , we get  $q = -7$ 

Hence, $pq=-14$