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Standard parametric equations of a parabola of the form $y^2=4ax$ are: $$ x(t)=at^2\\ y(t)=2at $$ which is fine since it can be easily verified. But is there any reason or advantage of making such a choice in the parametric equation of parabola ?

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    $\begingroup$ You've got $x(t), y(t)$ the wrong way round. $\endgroup$ – hypergeometric Jul 4 '17 at 16:11
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    $\begingroup$ Advantage compared to which other choice ? [By the way, I see nothing "standard" here.] $\endgroup$ – Yves Daoust Jul 4 '17 at 16:17
  • $\begingroup$ @YvesDaoust i understand we are free to make appropriate choice of parameters. but the above form is what usually used as the parametric form of the given parabola. So i was wondering is there any particular reason beyond simplicity of the equation? $\endgroup$ – ss1729 Jul 4 '17 at 16:35
  • $\begingroup$ @ss1729: I have rarely seen this form and you don't answer: what are the alternatives ? $\endgroup$ – Yves Daoust Jul 4 '17 at 16:40
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    $\begingroup$ IMO all forms $x=at,y=bt^2$ are virtually equivalent. I don't see an advantage of having $x=\dot y$. $\endgroup$ – Yves Daoust Jul 4 '17 at 17:14
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The standard Cartesian form equation for the parabola $y^2=4ax$ is significant because $a$ is the focal length, the focus of the parabola is $(a,0)$ and also because $4a$ is the length of the latus rectum.

For this parabola, the standard parametric equation $(at^2, 2at)$ is probably the simplest possible as it does not contain fractions. Other possibilities are $\left(\frac {t^2}{4a} , t\right), \left(\frac {t^2}a, 2t\right)$, which are not as neat.


Another example of a possible parametric equation is $\big(4a\sin t, 2a(1-\cos 2t)\big)$.

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  • $\begingroup$ thnx. i understand there s nothing standard in this particular form since we are free to choose the parameter as long as it satisfies the equation of the parabola. my doubt was avoiding fraction to make it simple is the only reason for such a choice ? $\endgroup$ – ss1729 Jul 4 '17 at 16:29
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    $\begingroup$ That's only my conjecture. Hopefully it's a logical one. $\endgroup$ – hypergeometric Jul 4 '17 at 16:32
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There is no standard parametrization.

A parametrization is beneficial if the parameter has an extra or particular geometrical or physical significance.

The given parametrization has focal length $a$. Differentiating $x$ wrt $y$ through $t,$ it can be appreciated that $t$ also represents tangent of angle which the tangent of parabola ( axis on $x$ axis) makes to the $y$ axis. It is also simple, algebraically.

EDIT1:

Another direct (unparametrized) oblique axes form with two branches with constants $ {(m,h,k)} $ is:

$$y= m x \pm \sqrt{m x h + k^2}$$

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