How the equation of a projectile represents a parabola? I am not able to prove that equation of motion of a projectile is parabola. The book simply says the  given below is the equation of a parabola but does not clarify it
$$y= {\tan\theta}x - \frac{g}{2(u\cos\theta)^2}x^2$$
But equation of parabola says
(i) $y^2=4ax$
(ii) $y^2=-4ax$
(iii) $x^2=4ay$
(iii) $x^2=-4ay$

None of the four options seem like the proper definition for a parabola. Mabe if you show that $y \sim x^2$ it would suffice.
Specifically, the vertex form of a parabola is $y = a (x-x_0)^2 + d$, where $a$, $x_0$ and $d$ are constants. Bring the given equation in this form to show it is indeed a parabola.

*

*Equate the two expressions $$  (\tan \theta) x - \frac{g}{4 v^2 \cos^2 \theta} x^2 = a \left(x - x_0 \right)^2 + d $$

*Foil the right-hand side to expand the terms $$  (\tan \theta) x - \frac{g}{4 v^2 \cos^2 \theta} x^2 = a x^2 -2 a x x_0 + a x_0^2 + d $$

*Match the coefficients of $x^2$, $x$ and the constant $$ \begin{aligned}
  - \frac{g}{4 v^2 \cos^2 \theta} x^2 &= a x^2 \\
  (\tan \theta) x &= -2 a x x_0 \\
  0 & = a x_0^2 + d
\end{aligned}$$

*Solve the three equations for $a$, $x_0$ and $d$ respectively $$ \begin{aligned}
  a & = -\frac{g}{4 v^2 \cos^2 \theta} \\
  x_0 & = \frac{2 v^2 \sin \theta \cos \theta}{g} \\
  d & = \frac{v^2 \sin^2 \theta}{g}
\end{aligned}$$

*Since all the terms above are constant and do not depend on $x$, the two expressions are indeed equal to each other, and we have the trajectory in "parabolic" form

$$y = \underbrace{ \left( -\frac{g}{4 v^2 \cos^2 \theta} \right)}_a 
 \left( x - \underbrace{\left( \frac{2 v^2 \sin \theta \cos \theta}{g} \right)}_{x_0} \right)^2 + \underbrace{ \left(\frac{v^2 \sin^2 \theta}{g}\right)}_d$$
