Convert this parametric parabola to general conic form

algebra-precalculusanalytic geometry

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?

Best Answer

$\lim_\limits{t\to \infty} \frac {y(t)}{x(t)}$ 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.

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

And in the general quadratic...

$Ax^2 + Bxy + Cy^2 + \cdots...$

$\frac {A-C}{B} = \cot 2\theta$ gives the rotation angle from standard....

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