The equation of a parabola with vertex at the origin is given by $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with the contraints, $B^2-4AC=0$ and $F=0$. There exists a parametrization as follows: $$r(t)=\left(\begin{array}{cc}at^2+bt\\ct^2+dt\end{array}\right)$$ I believe this is true because Geogebra will show me the numeric parametrization going either direction. That is, I can put in any $A,B,C,D,E$ or I can put in $a,b,c,d$ and the program will show me the other set. I just have to play by the constraint rules above. How do I find variables $a,\;b,\;c$ and $d$? What I have tried?
Brute force. Use a CAS and substitute $x=at^2+bt$ into the general conic form and solve for $y$. From the mess of terms, I collected the ones where $B^2-4AC$ could be factored away thus reducing the complexity. What remains is $$r(t)=\left(\begin{array}{cc}at^2+bt\\\frac{-B(at^2+bt)-E\pm\sqrt{(2BE-4CD)(at^2+bt)+E^2}}{2C} \end{array}\right)$$ This is a parametrization and it will half-graph but it isn't in the correct form. The $y$ term didn't become $ct^2+dt$. I did notice in passing that the root term is somehow related to the eigenvectors. It was elusive.
- Discovery using numerical examples showed me something. In the desired parametrization let the parabola's focal point be $(f_x,f_y),$, then $a=f_x$ and $b=2f_y$ and $c=f_y$ and $d=2f_x$. I can get the focal point by rotating the graph to the vertical with $\lambda x^2+E_1y=0$ (eigenval/vec rotation method) and using $x^2=4py$ to get p and then rotating back again.
I know that "discovery" is sort of an answer to my question, but I really want to know how to derive $a,b,c,d$ rather than my didactic method for finding them.