$$\begin{gathered} Given curve y^2=\alpha x^3-\beta \\ diff w.r.to x on boyh sides \\ 2y\frac{dy}{dx}=\alpha \left(3x^2\right) \\ \Rightarrow \frac{dx}{dy}=\frac{2y}{3\alpha x^2} \\ Slope of normal={\left(-\frac{\mathrm{dx}}{\mathrm{dy}}\right)}_{(2,3)}=-\frac{2\times 3}{3\mathrm{\alpha }{\left(2\right)}^2}=-\frac{1}{2\mathrm{\alpha }}=-\frac{1}{4} \\ \Rightarrow \mathrm{\alpha }=2 \\ \mathrm{Also}, (2,3) \mathrm{lies} \mathrm{on} \mathrm{the} \mathrm{curve}. \\ \Rightarrow 9=8\alpha -\beta \Rightarrow \beta =16-9=7 \\ Now \alpha +\beta =9 \end{gathered}$$