# How to convert this parametric parabola to general conic form?

How to convert parametric parabola to general conic form? Or, even better, how to find $$p$$ and $$θ$$ as new parameters. As part of a study for finding the vertex of a parabola, I made up a simple parametric parabola. $$\mathbf{r}:\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} 2t^{2}-2t+1\\ -2t^{2}+5t-1 \end{array}\right)$$ I was using it to find the vertex by minimizing the magnitude of the tangent vector. That worked OK and the vertex was found to be $$(h,k)=(25/32,59/32).\,$$ But then, I wanted to convert it to be parametrized as $$\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} h\\ k \end{array}\right)+\left(\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right)\left(\begin{array}{c} (2p)\tau\\ (p)\tau^{2} \end{array}\right)\tag{1}$$ I changed the equation parameter from t to τ because the two parametrizations are not the same.

From here I get a bit stuck. I tried to get $$θ$$ and $$p$$ by finding a couple of points $$(x,y)$$ on the parabola and I hoped to match coefficients - but there weren't any. Nor could I get enough information to solve for $$p$$ and $$θ$$. So then, I decided to convert it to general conic form, but oops - I didn't know how to do that either. Geogebra will just tell me the answer!. It is $$−2x^2−4xy−2y^2+15x+6y−9=0$$. I know how to rotate this and find $$θ$$ and $$p$$. I do not know how to convert $$\mathbf{r}$$ into the general conic? Both equations, when solved for $$t$$ give $$\pm$$parts and are unsuitable for substitution to get the general conic. So, How do it know?

• For the general $xy$ form ... (Assuming the "$=$" in your equation is a typo for "$+$") You have the system $x=2t^2-2t+1$, $y = -2t^2+5t-1$. Adding the equations gives $x+y=3t$. You can easily solve this for $t$ and substitute into either of the original equations. Done! – Blue Sep 23 at 23:17
• @Blue OK thank you. At least I can solve it all now. That part done. Is there any way to get p and $\theta$ without doing the rotation? as in staying only in a parametrized form? – Narlin Sep 23 at 23:26

$$\lim_\limits{t\to \infty} \frac {y(t)}{x(t)}$$ gives the slope of the axis of symmetry the parabola.

Since $$\lim_\limits{t\to \infty} \frac {y(t)}{x(t)} = \infty$$ is a parabola in standard position, our rotation angle $$\theta$$ is $$\frac \pi 2$$ off from standard.

$$\lim_\limits{t\to \infty} \frac {y(t)}{x(t)} = \tan (\theta+\frac {\pi}{2}) = -\cot\theta$$

$$Ax^2 + Bxy + Cy^2 + \cdots...$$
$$\frac {A-C}{B} = \cot 2\theta$$ gives the rotation angle from standard....