# What is the significance of the parameter in the parametric equation of a parabola?

The parametric equation of the parabola $$y^2=4ax$$ is given to be $$y=2at$$ and $$x=at^2$$ where $$t$$ is the parameter. What is the significance of the parameter $$t$$ in this equation?

While searching this website regarding this, I found the following statement from this answer to the question - On The Parametric Equation Of A Parabola for the variable point $$P(at^2,2at)$$:

• when $$t=0$$, $$P$$ is on the vertex
• when $$t>0$$, $$P$$ is on the upper branch
• when $$t<0$$, $$P$$ is on the lower branch

But this fact was pretty obvious.

Let me clarify what I mean by "significance". We know that the parametric equation of a straight line is given by: $$\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin\theta}=r$$ where $$r$$ is the parameter. Here, the "significance" of $$r$$ is - its magnitude is equal to the distance of any point $$(x,y)$$ on the line from the fixed point $$(x_1,y_1)$$ along the line.

Similarly, what does $$t$$ imply? Further, will the significance of $$t$$ depend on which type of parabola (opening right or left or upwards or downwards) it is referring to?

• In parabola parameter do not have any physical significance it's only a real number such that every point on parabola can be represented on parabola in terms of it and every distinct parameter will represent different point on parabola.for more on parabola you can also refer to mathsdiscussion.com/parabola Feb 2, 2020 at 5:02
• @Rajan: Thanks for the comment. May I know the reason for the absence of any significance to the parameter? Feb 2, 2020 at 5:05
• For this let's try to understand what is a parameter and why we need it. As far my understanding goes is parameter is an known selcected based on criteria that (1) in terms of which we can represent both x,y coordinate of all points on curve ( There should not be any point on curve which cannot be represented by that known.(2) Every different parameter should represent different point . And no two different parameter represent same point. And use is in any calculation instead of two unknowns x,y we have only one. Feb 2, 2020 at 5:11
• @Rajan: I'm sorry. Even though I'm well aware of these facts, I'm unable to say why there must be no significance. I think it's better to answer using the answer box below instead of using the comment section. Feb 2, 2020 at 5:15
• Also in case of line inclination of every point on line with reference / fixed point is same and distance changes for all point hence we take distance as parameter , in case of circle distance of all point remain same hence we take inclination of radius vector with +ve x-axis as parameter. where as in case of parabola both distance and inclination changes hence we select any real number is taken as parameter. Feb 2, 2020 at 5:16

Here's a picture of the situation (rotated by $$90^\circ$$), from which, as a bonus, we may deduce the parabola-ness of the curve from the parameterization (without relying on "knowing" it corresponds to $$y=\frac{1}{4a}x^2$$): Point $$F$$ (the parabola's ostensible focus) is $$a$$ units "up" from origin $$O$$. Point $$P$$ is parameterized by $$P=(2at,at^2)$$, so $$|OP'|=2at$$ and $$|PP'|=at^2$$. Let $$M$$ be the midpoint of $$\overline{OP'}$$, and let $$D$$ be the point where the extension of $$\overline{PP'}$$ meets the ostensible directrix $$a$$ units "down" from $$O$$.

Clearly, $$\triangle OFM\cong\triangle P'DM$$. Also, $$\triangle OFM\sim\triangle P'MP$$ by the key proportionality of corresponding sides $$\frac{|OM|}{|OF|}=\frac{at}{a}=t=\frac{at^2}{at}=\frac{|PP'|}{|MP'|} \tag{1}$$ But, then $$\triangle MFP$$ must be a right triangle whose legs are in the same proportion $$\frac{|MP|}{|MF|}=t \tag{2}$$ so that $$\angle MFP\cong\angle OFM\cong\angle MDP$$, and we deduce $$\overline{PF}\cong\overline{PD}$$: therefore, $$P$$ lies on the parabola with focus $$F$$ and directrix through $$D$$. $$\square$$

To the question at hand ... By the Reflection Property of Parabolas, the bisector of $$\angle FPD$$ is tangent to the parabola at $$P$$. Thus, $$\overleftrightarrow{MP}$$ is a tangent line, and right-hand half $$(1)$$ shows that $$t$$ is its slope. (This is consistent with @mathlove's comment and @ChiefVS's answer that, in the original horizontally-opening context, the slope of the tangent is $$1/t$$.)

Of course, the left-hand half of $$(1)$$, as well as $$(2)$$, give other interpretations of $$t$$.

• Thank you for your answer. A minor issue: May I know the reason for using "$\cong$" over "$=$" for the equality of angles below equation $(2)$ (so far I have used it only for congruence)? Further, which GeoGebra platform do you use to make such beautiful diagrams - Graphing calculator, Geometry or GeoGebra classic? Feb 13, 2020 at 14:47
• @GuruVishnu: Formally, "$\cong$" is for objects (segments, angles, triangles, etc) and "$=$" is for values (lengths, measures, areas, etc). So, I write "$\angle A\cong\angle B$", but "$m\angle A=m\angle B$". ... Actually, with angles, I tend to be a little informal, because the "$m$" looks like multiplication, so I'm perfectly happy writing things like "$\angle A=90^\circ$" and "$\angle A=2\angle B$" and "$\angle A=\angle B-\angle C$". Even so, when comparing one angle to another directly, I tend to favor "$\angle A\cong\angle B$" over "$\angle A=\angle B$".
– Blue
Feb 13, 2020 at 15:46
• @GuruVishnu: I use the desktop GeoGebra Classic (5.0.x) for the Mac. ... As an app maker and user interface stickler, I'm not terribly fond of the Calculator and Geometry apps on the desktop; they are more "modern", but they look and act like wrappers for their web-apps. That's efficient for their developers, but unsatisfying for their users like me. Even the cross-platform, Java-based Classic is annoyingly non-native, but I'm stuck with it ... until I write my own alternative. ;) ... I complain, but the app helps me make "beautiful diagrams" (thanks for that! :), so I'm pretty happy.
– Blue
Feb 13, 2020 at 16:07

Geometric Interpretation of the parameter $$t$$ is that $$t^{-1}$$ is the slope of tangent at point $$P(at^2,2at)$$ on the parabola $$y^2 = 4ax$$. It can be easily found, so I'm hinting how to get it:

Since the equation of tangent at any point $$(x_0,y_0)$$ of any general curve is

$$(y-y_0)=(y')_{(x_o,y_0)}(x-x_0)$$.

In the case of standard parabola, it becomes $$ty = x + at^2$$.

Differentiate $$y$$ wrt $$x$$ via parameter $$t$$ by Chain Rule

$$\frac{(2at)'}{(a t^2)'} = \frac{2a}{2at} = \frac{1}{t}$$

So $$t$$ is interpreted as the negative slope of tangent angle clockwise to parabola at any point, to $$y$$ axis at any chosen or its $$t$$.

The parabola is $$y^2= 4 a x$$ is obtained by eliminating the parameter $$t$$ between $$y(t), x(t)$$ of course after understanding its geometrical significance.