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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?

  1. 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.

    1. 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.

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marked as duplicate by Blue geometry Nov 12 '18 at 11:34

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  • $\begingroup$ Eliminating $t$ from your parametrization, I get $c^2x^2-2acxy+a^2y^2+(bcd-ad^2)x+(-b^2c+abd)y$. $\endgroup$ – Jan-Magnus Økland Nov 11 '18 at 19:42
  • $\begingroup$ My attempt to eliminate $t$ did not get your equation and I wasn't able to simplify what I got. Also, if I try to use your equation, it suggests that $\sqrt{A}=c$ and that D and E are equal to the terms in front of $x$ and $y$, however, with an actual example, either my A term was negative, thus $c=\sqrt{A}$ doesn't work, or I can multiply all thru by -1 and then the D term, $D=(bcd-ad^2)$ fails to solve with a negative root. Maybe if you show some steps in getting t eliminated. $\endgroup$ – Narlin Nov 11 '18 at 20:36
  • $\begingroup$ In M2 R=QQ[a,b,c,d] S=R[t,x,y,MonomialOrder=>Lex] I=ideal(x-a*t^2-b*t,y-c*t^2-d*t) gens gb I $\endgroup$ – Jan-Magnus Økland Nov 11 '18 at 20:42
  • $\begingroup$ Might be a little over my head, but I'll work on it. I do see what you did. $\endgroup$ – Narlin Nov 11 '18 at 20:47
  • $\begingroup$ The coefficients I get with the $t$ elimination equation work all the time when the parabola opens anywhere in quadrant 1. In quadrants 3 or 4, the $\sqrt(negative num)$ comes up all the time. In quadrant 2, the coefficients yield a parabola rotated 90 deg into quadrant 1. M2 doesn't really show me how they simplified the equation. $\endgroup$ – Narlin Nov 11 '18 at 21:28