A particle moves in the xy pane with a constant acceleration 'g’ in the negative y-direction. Its equation of motion is $y=ax-b{x}^2,$  where a and b are constants. Which of the following are correct?

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

Given $y=ax-b{x}^2$

On comparing $\tan \theta =a$ and $\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b=\frac{g}{2u^2{\cos }^2\theta }$

$\therefore 1+{\tan }^2\theta ={\sec }^2\theta =1+a^2$

vertical velocity is given by $u\sin \theta$, on substituting the value in $a\sqrt{\frac{g}{2b}}$we get $u\sin \theta$.

As we know the x-component of the velocity will be contstant.