Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola ?

When a body is projected into space at an angle θ (other than 90°) with the horizontal is called projectile.
To prove that the trajectory of a projectile is a parabol a :

Consider a body projected into air with an initial velocity ’u’ making an angle $\mathrm{\theta }$ with the horizontal. Initial velocity ’u’ can be resolved into two rectangular components. Horizontal component u_x = ucos $\mathrm{\theta }$  and vertical component u_y = u sin$\mathrm{\theta }$ .Along horizontal direction there is no acceleration hence the projectile move with uniform velocity along horizontal direction.
Equation for the trajectory of the projectile :
Let the projectile be at the point p(x, y) after time t. x and y are the horizontal and vertical displacements after time t.
Along horizontal direction :
Intial velocity (u_x) = ucos$\mathrm{\theta }$  acceleration
(a_x) = 0 displacement (S_x) = x
Time (t) = t
From
$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\Rightarrow {\mathrm{S}}_{\mathrm{x}}={\mathrm{u}}_{\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2$
$\mathrm{x}=(\mathrm{ucos}\mathrm{\theta })\mathrm{t} .....\left(1\right) \mathrm{t}=\frac{\mathrm{x}}{\mathrm{ucos}\mathrm{\theta }}.....\left(2\right)$
Along vertical direction
Initial velocity $={\mathrm{u}}_{\mathrm{y}}=\mathrm{usin}\mathrm{\theta }$
acceleration ${\mathrm{a}}_{\mathrm{y}}=-\mathrm{g}$
Displacement travelled $\left({\mathrm{s}}_{\mathrm{y}}\right)=\mathrm{y}$
Time (t) = t
$\begin{array}{l}\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\dots .\left(3\right){\mathrm{S}}_{\mathrm{y}}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2\\ \mathrm{y}=\mathrm{u}(\sin \mathrm{\theta })\left(\frac{\mathrm{x}}{\mathrm{ucos}\mathrm{\theta }}\right)-\frac{1}{2}\mathrm{g}\left(\frac{{\mathrm{x}}^2}{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}\right)\\ \mathrm{y}=\tan \mathrm{\theta x}-\left(\frac{\mathrm{g}}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}\right){\mathrm{x}}^2\end{array}$
Let $\tan \mathrm{\theta }=\mathrm{A} , \left(\frac{\mathrm{g}}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}\right)=\mathrm{B}$
$\therefore \mathrm{y}=\mathrm{Ax}-{\mathrm{Bx}}^2(\because \mathrm{A}\text{ and }\mathrm{B}\text{ are constants })$
This is the equation of the trajectory of the projectile