If x + 4y = 14 is a normal to the curve y² = $\mathrm{\alpha }$x³ - $\mathrm{\beta }$ at (2,3) then the value of $\mathrm{\alpha }+\mathrm{\beta }$ is___

${\mathrm{y}}^2={\mathrm{\alpha x}}^3-\mathrm{\beta }\text{ or }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3{\mathrm{\alpha x}}^2}{2\mathrm{y}}$
Therefore, slope of the normal at (2,3) is 
${\left(-\frac{\mathrm{dx}}{\mathrm{dy}}\right)}_{(2,3)}=-\frac{2\times 3}{3\mathrm{\alpha }(2{)}^2}=-\frac{1}{2\mathrm{\alpha }}=-\frac{1}{4}$
or $\mathrm{\alpha }$=2 
Also, (2,3) ;lies on the curve. Therefore,
$9=8\mathrm{\alpha }-\mathrm{\beta }\text{ or }\mathrm{\beta }=16-9=7\text{ or }\mathrm{\alpha }+\mathrm{\beta }=9$