Parameterization of the parabola

It is (well?) known that if \(x = at^2+bt+c\) and \(y=pt^2+qt+r\), then

\[ (Ax+By)^2+Cx+Dy+E=0 \]

where

\[\begin{align*} A&=p\\ B&=-a\\ C&=qv_2-2pv_1\\ D&=-bv_2+2av_1\\ E&=v_1^2-v_2v_3 \end{align*}\]

with \(\langle v1, v2, v3\rangle =\langle a,b,c\rangle \times \langle p,q,r\rangle\); that is, the \(v_i\) values are the elements of the cross product of the vectors of the coefficients.

In other words, two quadratic functions parameterize a parabola.

Finding the equations which parameterise a given parabola

Suppose we have a general parabola given by \((Ax+By)^2+Cx+Dy+E=0\). Clearly we need to choose coefficients of \(t^2\) in \(x\) and \(y\) in such a way that they cancel out in the first bracket.

We can start with, say

\[\begin{align*} x&=Bt^2+bt+c\\ y&=-(At^2+qt+r) \end{align*}\]

Then

\[\begin{align*} (Ax+By)^2&=((ABt^2+Abt+Ac)-(BAt^2+Bqt+Br))^2\\ &=((Ab-Bq)t+(Ac-Br))^2\\ &=(Ab-Bq)t^2+2(Ab-Bq)(Ac-Br)t+(Ac-Br)^2 \end{align*}\]

Adding this to the rest of the expression (Cx+Dy+E) and collecting "like terms", we end up with:

\[\begin{align*} t^2:&\quad (Ab-Bq)^2+CB-DA=0\\ t:&\quad 2(Ab-Bq)(Ac-Br)+Cb-Dq = 0\\ 1:&\quad (Ac-Br)^2+Cc-Dr+E=0 \end{align*}\]

From the first equation, if \(b=s\), say, then

\[ q = \frac{\sqrt{DA-CB}-As}{B}. \]

But we don't want square roots if we can avoid them. One thing we can do is to introduce an extra multiplicative variable into the original equations:

\[\begin{align*} x&=k(Bt^2+bt+c)\\ y&=-k(At^2+qt+r) \end{align*}\]

Then the three equations corresponding to the coefficients \(t^2,t,1\) can be written as

\[\begin{align*} &(Ab+Bq)^2k^2+(BC-AD)k=0\\ &2(A b + B q)(A c + B r)k^2 + (Cb+Dq)k=0\\ &(Ac+Br)^2k^2+(Cc+Dr)k+E=0 \end{align*}\]

The first equation can be solved to produce two solutions:

\[ b=s,\quad q = \frac{-Aks+\sqrt{(AD-BC)k}}{Bk} \]

and

\[ b=t,\quad q = \frac{-Akt-\sqrt{(AD-BC)k}}{Bk} \]

Clearly to eliminate the square root we can set \(k=1/(AD-BC)\). This produces the solutions

\[ b=s,\quad q = \frac{-As-(AD-BC)}{B} \]

and

\[ b=t,\quad q = \frac{-At+(AD-BC)}{B} \]

Since \(s\) and \(t\) are arbitrary values, and we only want one solution to our equations, we can choose \(s=-D\) and \(t=D\). Then both solutions collapse to:

\[ b=D,\quad q = \pm C. \]

In fact, we can do the entire solution using a computer algebra system (in this case SageMath). We start by creating the variables we need, finishing with a polynomial in \(t\):

<Sage>: var('A,B,C,D,E,x,y,t,b,c,q,r')
<Sage>: x = k*(B*t^2 + b*t +c)
<Sage>: y = -k*(A*t^2 + q*t + r)
<Sage>: p = (A*x + B*y)^2 + C*x + D*y + E
<Sage:> pc = p.expand().poly(t)

The equations are the coefficients of this polynomial, which we require to be equal to zero:

<Sage>: eqns = [pc.coefficient(t,i).subs({k:1/(A*D-B*C)}).factor().numerator() for i in range(3)]
<Sage>: sols = solve(eqs,[b,c,q,r],solution_dict=True)

Now we substitute in D for the free variable given by \(p\), and simplify:

<Sage>: s = sols[0][b].free_variables()[0]
<Sage>: [sols[0][z].subs({s:-D}).full_simplify() for z in [b,c,q,r]]

This produces the output

\[ [-D, BE, -C, AE] \]

which means that the general parabola \((Ax+By)^2+Cx+Dy+E=0\) can be parameterized by

\[ x = \frac{Bt^2 - Dt +BE}{AD-BC},\qquad y = \frac{-At^2 + Ct -AE}{AD-BC} \]