The equation of a projectile is $y=px-q{x}^2.$ Then its horizontal range is

The equation for the trajectory of a projectile is $y=\tan \theta x-\frac{g}{2u^2{\cos }^2\theta }{x}^2$  
$y=x\tan \theta (1-\frac{gx}{u^{2}2\tan \theta {\cos }^2\theta })=x\tan \theta (1-\frac{gx}{u^2\sin 2\theta })=x\tan \theta (1-\frac{x}{R})$ (R is range)
If the given equation is converted into this form $y=px-q{x}^2=px(1-\frac{qx}{p})$
On comparison we get $R=\frac{p}{q},$ i.e. horizontal range is $\frac{P}{q}$  
So the correct answer is 4