# How can we fit hyperbola to data?

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ishita agrawal le 10 Sep 2017
Commenté : Star Strider le 30 Sep 2022
Hi, I have x and y coordinates for my data points. The data is fitted well with exponential fitting. However, I have to fit a hyperbola (a must condition for my results). I am using code,
% fit data k=0.002; % hit and trial
x = 1/xdata; y = k*1/x; p=plot(x,y)
I have attached excel file of my data. Accept my thanks in advance.
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### Réponse acceptée

Star Strider le 10 Sep 2017
Try this:
D = xlsread('data for hyperbola fitting.csv');
D = sortrows(D,1);
x = D(:,1);
y = D(:,2);
hyprb = @(b,x) b(1) + b(2)./(x + b(3)); % Generalised Hyperbola
NRCF = @(b) norm(y - hyprb(b,x)); % Residual Norm Cost Function
B0 = [1; 1; 1];
B = fminsearch(NRCF, B0); % Estimate Parameters
figure(1)
plot(x, y, 'pg')
hold on
plot(x, hyprb(B,x), '-r')
hold off
grid
text(0.7, 0.52, sprintf('y = %.4f %+.4f/(x %+.4f)', B))
‘Accept my thanks in advance.’
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Rik le 13 Mar 2021
Have you read the Wikipedia article about the R2? It is actually fairly easy to write code that calculates it.
Star Strider le 30 Sep 2022
D = sortrows(D,1);
x = D(:,1);
y = D(:,2);
hyprb = @(b,x) b(1) + b(2)./(x + b(3)); % Generalised Hyperbola
B0 = [1; 1; 1];
mdl = fitnlm(x, y, hyprb, B0)
mdl =
Nonlinear regression model: y ~ b1 + b2/(x + b3) Estimated Coefficients: Estimate SE tStat pValue __________ ________ ________ __________ b1 0.094879 0.015709 6.0399 2.4472e-09 b2 0.14875 0.017335 8.5807 5.5591e-17 b3 -0.0097497 0.029435 -0.33123 0.74057 Number of observations: 739, Error degrees of freedom: 736 Root Mean Squared Error: 0.0502 R-Squared: 0.772, Adjusted R-Squared 0.772 F-statistic vs. constant model: 1.25e+03, p-value = 2.94e-237
B = mdl.Coefficients.Estimate;
figure(1)
plot(x, y, 'pg')
hold on
plot(x, hyprb(B,x), '-r')
hold off
grid
xlabel('x')
ylabel('y')
text(0.7, 0.52, sprintf('$y = %.4f + \\frac{%.4f}{x %+.4f}$', B), 'Interpreter','latex', 'FontSize',12)
.

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