If the normal form of the equation of a straight line 4x+ 3y +2=0 is $x\cos \alpha +y\sin \alpha =p$ and its intercept form is $\frac{x}{a}+\frac{y}{b}=1,$ then $\frac{p\sec \alpha }{ab}=\mathrm{.........}$

Normal form:Given equation is $4x+3y+2=0\Rightarrow -4x-3y=2\Rightarrow -\frac{4}{5}x-\frac{3}{5}y=\frac{2}{5}$

$\Rightarrow \cos \alpha =-4/5,\sin \alpha =-3/5,p=2/5.$

 Intercept form:  Given equation is $4x+3y+2=0\Rightarrow 4x+3y=-2\Rightarrow \frac{4x}{-2}+\frac{3y}{-2}=1\Rightarrow \frac{x}{-1/2}+\frac{y}{-2/3}=1$

$\Rightarrow a=-\frac{1}{2},b=-\frac{2}{3}\Rightarrow \frac{p\sec \alpha }{ab}=\frac{\frac{2}{5}\left(-\frac{5}{4}\right)}{\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)}=\frac{-1/2}{1/3}=-\frac{3}{2}$