 # I have a curve in 2D space (x,y), described by the following equation: ax^2+bxy+cy^2+d=0 where a,b,c,d are known. It is obvious that it is a 1D curve embedded in a 2D space. So I would think there could be such a description of the curve, where only single parameter is present. enmobladatn 2022-07-18 Answered
2D curve with two parameters to single parameter
I have been thinking about the following problem. I have a curve in 2D space (x,y), described by the following equation: $a{x}^{2}+bxy+c{y}^{2}+d=0$ where a,b,c,d are known. It is obvious that it is a 1D curve embedded in a 2D space. So I would think there could be such a description of the curve, where only single parameter is present.
It is obvious that you can plug x and then solve a quadratic equation for y, but that is not what I'm looking for. The solution I expect is in the form $x={f}_{x}\left(t\right),y={f}_{y}\left(t\right),t\in ?$
which can be used in the case of circle equation with sine and cosine of angle.
My goal is to plot the curve in python, so I would like to start at some point of the curve and trace along it. Could you point me to a solution or some materials which are dedicated for such problems?
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Explanation:
Assuming $a>0$ and $c>0$, why not to use $x=\frac{r\mathrm{cos}\left(t\right)}{\sqrt{a}}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}y=\frac{r\mathrm{sin}\left(t\right)}{\sqrt{c}}$ which make $a{x}^{2}+bxy+c{y}^{2}+d$ to become ${r}^{2}\left(1+\frac{b\mathrm{sin}\left(2t\right)}{2\sqrt{ac}}\right)+d=0$ then r that you replace now in x and y.

We have step-by-step solutions for your answer! jlo2ni5x
Step 1
The solution of the quadratic equation gives you one kind of parameterization of the ellipse, though you may not like it. You could find the minimum and maximum values of x and let x go from the minimum to the maximum and back again to the minimum as t increases. Going in one direction you set y to one of the solutions of the quadratic, and in the other direction you set y to the other solution, so that y is also a function of t.
There are other ways, however.
Since the equation is a quadratic equation in x and y, there are only a limited number of kinds of shapes that can solve it, and we can quickly rule out all of these shapes except for an ellipse. It is possible to find the center, major axis a, and minor axis b of the ellipse, as well as the angle the axes are rotated from the coordinate axes. Let ${x}^{\prime }=a\mathrm{cos}\theta$ and ${y}^{\prime }=b\mathrm{sin}\theta$ to get an axis-aligned ellipse centered at (0,0), the apply a coordinate transformation $x=f\left({x}^{\prime },{y}^{\prime }\right),$ $y=g\left({x}^{\prime },{y}^{\prime }\right)$ to translate and rotate the ellipse as needed. (The translation is zero in this case since it turns out the center of your ellipse is at (0,0).)
Step 2
You can save a lot of bother from the previous paragraph by knowing that at the end of the procedure you'll have something equivalent to $\begin{array}{rl}x& =h+m\mathrm{cos}\theta -n\mathrm{sin}\theta ,\\ y& =k+n\mathrm{cos}\theta +m\mathrm{sin}\theta .\end{array}$
Take a few points on the ellipse and solve for h, k, m, and n. (In this case $h=k=0,$, but I've written this for a general ellipse.)

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