Prove that projectile motion is parabola?

Path of a projectile
let OX be a horizontal line on the ground and OY be a verticle line O is the origin for X and Y axis Consider that the projectile is fired with velocity u and making angle θ with the horizontal form the point ′O′ on the ground
The velocity of projection of the projectile can be resolved into the following two components 
(i) u_x = ucosθ, alongOX
(ii) u_y = usinθ, alongOY
 As the projectile moves it covers distance along the horizontal due to the horizontal component ucosθ of the velocity of projection and along verticle due to the vertical component usinθ . let that any time t , the projectile reaches the point P , So that its distance along the X and Y axis are given by x and y respetively.
Motion along horizontal direction : it we neglect friction due to air, then horizontal component of the velocity i.e. ucosθ will remain constant. thus
Initial velocity along the horizontal u_x = ucosθ
Acceleration along the horizontal a_x = 0
The position of the projectile along X-axis at any time t is given by
x = u_xt + $\frac 12$a_xt²
Putting ux ​= ucosθ and ax = 0 we have 
x = (u cos θ)t + $\frac 12$​(o)t²
or x = (u cos θ)t
or t = $\frac x{u \cos \theta}$ ​....... (i)
Motion along verticle direction: 
The velocity of the projectile along the verticle goes on decreasing due to effect of gravity initial velocity along vertical u_y = usinθ
Acceleration along vertical a_y=−g
The position of the projectile along T-axis at any time t is given by 
y = u_yt + $\frac 12$a_yt²
Putting uy​=usinθ and ay​=−g we have 
y = (u sin θ)t + $\frac 12$(−g)t²
or y = (u sin θ)t − $\frac 12$gt².....(2)
Putting the value of t from equation (1) in equation (2) we get 
y = (u sin θ) $\frac x{u \cos \theta}$ − $\frac 12$ ​g($\frac x{u \cos \theta}$)²
or y = x tan θ − ($\frac g{2\ u^2\ cos^2\theta}$)x²...(3)
This is an equation of a parabola . hence the path of projectile projected at some angle with the horizontal direction is a parabola.