Characteristics of parallel parabola (offset curve) and the formula to find the equation

Question

What are the characteristics of parallel parabola? And is there any formula to find the equation of parallel parabola if we know the equation of one parabola?

Answer 1

(see graphics below obtained with Geogebra, using formulas in the text).

Preliminary remark: The keyword "parallel curve" is ambiguous ; you see it replaced by "offset curve" in many works.

If a curve is defined parametrically by $X=\binom{x=x(t)}{y=y(t)}$, a tangent vector is $T=\binom{x'(t)}{y'(t)}$ (speed vector); we can take as normal vector $N=\binom{ \ \ y'(t)}{-x'(t)}$ (check that the dot product $N \bullet T$ is 0), and the unit norm normal vector is $N/\|N\|=\binom{y'(t)/\sqrt{x'^2(t)+y'^2(t)}}{-x'(t)/\sqrt{x'^2(t)+y'^2(t)}},$ with a certain orientation (outwards or inwards with respect to the initial curve.)

Thus, if you want to be at distance $a$ from the initial curve, you proceed from the generic point by making a step of length $a$ in the direction of vector $N/\|N\|$ giving the parametric representation that can found as well in the Wikipedia article on parallel and offset curves:

$$\tag{1}\left\{\begin{aligned}X&=x+a{\frac {y'}{\sqrt {{x'}^{2}+{y'}^{2}}}}\\Y&=y-a{\frac {x'}{\sqrt {{x'}^{2}+{y'}^{2}}}}.\end{aligned}\right.$$

Let us take the case of the standard parabola $y=x^2$ with parametrization $x=t, \ y=t^2$; we get: $x'=1, \ y'=2t$, yielding

$$\tag{2}\left\{\begin{aligned}X&=t+\frac {2at}{\sqrt {1+4t^2}}\\Y&=t^2-{\frac {a}{\sqrt {1+4t^2}}}.\end{aligned}\right.$$

which is a parametric representation that allows to draw any parallel curve to the initial parabola. According to the sign of $a$, these curves will be external or internal to the parabola (I understand you are interested in external curves).

One can, using a CAS (Mathematica in my case), eliminate $t$ between the 2 expressions above ; one gets a 12th degree implicit equation that can be expressed as the product of two 6th degree factors in variables $X,Y$:

$$\left( 9\,a^2 - 24\,a^4 + 16\,a^6 - 92\,a^2\,X^2 - 48\,a^4\,X^2 - 9\,X^4 + 48\,a^2\,X^4 - 16\,X^6 + 24\,a^2\,Y - 32\,a^4\,Y + 18\,X^2\,Y + 24\,a^2\,X^2\,Y + 8\,X^4\,Y - 9\,Y^2 + 40\,a^2\,Y^2 - 16\,a^4\,Y^2 + 32\,X^2\,Y^2 + 32\,a^2\,X^2\,Y^2 - 16\,X^4\,Y^2 - 24\,Y^3 + 32\,a^2\,Y^3 + 32\,X^2\,Y^3 - 16\,Y^4 \right) \, \left( a^2 + 8\,a^4 + 16\,a^6 + 20\,a^2\,X^2 - 48\,a^4\,X^2 - X^4 + 48\,a^2\,X^4 - 16\,X^6 - 8\,a^2\,Y - 32\,a^4\,Y + 2\,X^2\,Y - 8\,a^2\,X^2\,Y + 40\,X^4\,Y - Y^2 + 8\,a^2\,Y^2 - 16\,a^4\,Y^2 - 32\,X^2\,Y^2 + 32\,a^2\,X^2\,Y^2 - 16\,X^4\,Y^2 + 8\,Y^3 + 32\,a^2\,Y^3 + 32\,X^2\,Y^3 - 16\,Y^4 \right)=0$$

(please note that $a$ is always present with an even exponent).

One of the factors accounts for the external curve at distance $a$, the other for the internal curve at distance $a$. These expressions are untractable for practical applications.

These offset curves being polynomials with degree 6, they aren't conic curves (2nd degree polynomials), even if they look for example like parabolas when they are close to the "mother curve" (but with an increasingly circular aspect at the bottom).

It is to be noted that internal parallel curves can have a spike.

The second figure gives a different view: the offset curves are implicitly realized as "envelopes" of circles (think to the parabola as an atoll and the inside of the circles as "territorial waters" within a same distance).

Remark: As @WimC remarked, one can separate the inside and outside offset curves, giving the example of parabola $x=y^2$ for which the inside offset curve at distance $1$ has the following 6th degree equation $$16 x^4 - 32 x^3 y^2 - 40 x^3 + 16 x^2 y^4 + 9 x^2 - 40 x y^4 + 6 x y^2 +40 x + 16 y^6 - 47 y^4 + 28 y^2 - 25 = 0 $$

Answer 2

This might answer your question.

In the case of:

$y = a x^2$

Where:

$x$ = coordinate for point on the Parabola

$y$ = coordinate for point on the Parabola

$a$ = dilation coefficient of the Parabola

$X_d$ = coordinate for point displaced normal to a tangent of the Parabola

$Y_d$ = coordinate for point displaced normal to a tangent of the Parabola

$d$ = displaced amount

Then:

$X_d = x ± \frac{2dx}{2(\sqrt{.25a^2+x^2})}$

$Y_d = y ± \frac{.5da}{\sqrt{.25a^2+x^2}}$

Answer 3

Say you have a parabola of the form $ax^2+bx+c$.

Another parabola is parallel if it is of the form $ax^2+bx+k$, where $k$ is a constant that is not equal to $c$.

If $k=c$, then the parabolas would be identical.