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

Question

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?

Answer 1

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 $D$. $\square$

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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 $t$.

Answer 2

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 $ty = x + at^2$.

Answer 3

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.