I am trying to quantify the correlation of a given set of points to the set of points defined by a parametric equation, as well as find the best $l$ to fit the points. However, I am unsure of how to go about doing this.
The parametric equation I am working with is $$x(a)=a\ln \left(\frac{l+\sqrt{l^2+a^2}}{a}\right)$$ $$y(a)=\sqrt{l^2+a^2}-a.$$
I have found $l\approx.373$. The range of the parameter $a$ is $0\lt a \lt \infty$.
The set of points I am trying to analyze is \begin{array}{c|c} \style{font-family:inherit}{x} & \style{font-family:inherit}{\text{y}}\\\hline .372 & .017 \\\hline .343 & .125 \\\hline .308 & .184 \\\hline .267 & .232 \\\hline .234 & .262 \\\hline .192 & .293 \\\hline .147 & .320 \\\hline .110 & .338 \\\hline .065 & .356 \\\hline .030 & .367 \\\hline .003 & .372 \end{array}
A plot of the parametric equation and the set of points suggests a strong correlation for $l\approx.373$. Here is a graph on Desmos showing this: parametric equation and plotted points (The parametric equation is in red and the data points are plotted in black).
Typically, when finding the correlation between a function and a plot of points, the plot is linearized as the function indicates and a linear regression is run to evaluate the correlation. Unfortunately, I have found no way to linearize the set of points. Are there any other ways to find the parametric curve of best fit?
Plotting the data, just as you did, reveals the strong relation between $x$ and $y$.
Now, what I suspect is that they want you to find the best value of $l$ to improve the fit.
It is impossible to get rid of the parametrization in order to have $y=f(x)$. However, you can do the opposite.
Eliminate $a$ from $y$ $$a=\frac{l^2-y^2}{2 y}\qquad \implies \qquad x=\frac{(l^2-y^2) }{2 y}\,\log \left(\frac{l+y}{l-y}\right)$$
Using the starting value you gave, a basic nonlinear regression gives $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ l & 0.372631 & 0.000115 & \{0.37237,0.372892\} \\ \end{array}$$ and you can notice that $0.373$ is not in the confidence interval (even if very close to it).
For sure, this introduces a bias in the problem since what has been measured is $y$ and not $x$.
To compensate, I think that I should use orthogonal regression (total least squares).