Parametric Equation of parabola?

Question

I know that the cartesian equation of a parabola is y²=4ax. I also know that the parametric equation of a parabola is x = t², y = 2t but can someone please explain to me how you can get from the cartesian equation to the parametric equation.

Answer 1

u got it wrong
x=at^2 and y=2at

parabola y^2=4ax
=> y^2 / 4a = x
=> y^2 / 4a^2 = x/a
let this b equal to t^2

thus
y^2 / 4 a^2 = x/ a = t^2
thus y= 2at and x=at^2

Answer 2

In the first place there are many parametric equations for the parabola you mention. Your parametric equations are for a = 1.

You have to pay attention to the range of the variables that you want to parametrize over, in your example x >=0 and y is between + infinity and - infinity as is t.

But given this restriction you can make x (or y ) pretty much anything you want, say x = f(t), and then solve for y = 2 sqrtf(t)
or y = g(t) and x = 1/2(g(t))^2

Answer 3

If y^2=4ax then y= (4ax)^(1/2)=2b(x^(1/2)). The constant b is the square root of a.
Use the x-term to get t. Now y=2t and t=b(x^(1/2)), so t/b=x^(1/2).
So square both sides and x=(t^2)/a
y=2t
x=(t^2)/a
There’s your parametric equation.

Answer 4

get over the t:
x=t*t=>t=sqrt(x)
y=2t=>y=2sqrt(x)=>y*y=4x