Parabola from 4 approximate points

Question

I have calculated four approximate points from a sensors to get information. I would like to deduce the closest parabola to my points. The problem is that I can't solve it to get an appropriate result.

Here are my four points: $(414, 960), (1020,2340), (1387,3300), (1848,4510)$

Also, I tried to solve it with Wolfram Alpha using this instruction: solve $$ \begin{align*} 171396a+397440b+921600c+414d+960e+f&=0\\ 1040400a+2386800b+5475600c+1020d+2340e+f&=0\\ 1923769a+4577100b+10890000c+1387d+3300e+f&=0\\ 3415104a+8334480b+20340100c+1848d+4510e+f&=0\\ b^2=4ac \end{align*} $$

I wasn't able to make it works... I might not have the correct syntax. Any help will be appreciate.

Answer 1

In fact there are two parabolas in the $x-y$ plane that pass through your four points (exactly).

$$\eqalign{&103297015086160900\,{x}^{2}\cr + &\left( -78528720845214360+8356361780\, \sqrt {31761606515} \right) xy\cr + &\left( 14930193933347471-3176347356\, \sqrt {31761606515} \right) {y}^{2}\cr +& \left( -18003514867830528700- 7272001822800\,\sqrt {31761606515} \right) x\cr +& \left( 1629063111000\, \sqrt {31761606515}+6793046116045210090 \right) y\cr +& 678223769919769543800+1052877465525600\,\sqrt {31761606515} =0\cr} $$ and the same with $\sqrt {31761606515}$ replaced by $-\sqrt {31761606515}$

Answer 2

It seems you are trying to solve a problem which has no solution. Why not try a quadratic regression instead ?

EDIT: This answer is a good one, and Wolfram Alpha has a quadratic regression function you could try--just check the Examples.

Answer 3

The problem can be solved, as pointed by jbeuh, with a quadratic regression. I found this tool online that did the work for me (sorry, I am a lazy guy). The equation found is:

y = 2.043800872·10-4 x2 + 2.02022332 x + 84.68775484

Which gives a very good approximation of the equation.

Answer 4

Your problem is closely related to the fact that five points determine a conic section. To understand the tecnical reasons, look at this and this.

Here you can calculate your coefficients, and Geogebra has a nice surprise for you (it shows you how magic is done with the "Conic through five points tool").

Since you give only four points, it is not surprising that there are more than one conic that satisfy your condition, as Robert pointed (pun unintended).