# How can we fit hyperbola to data?

173 views (last 30 days)
ishita agrawal on 10 Sep 2017
Commented: Star Strider on 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.

Star Strider on 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))
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Star Strider on 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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