How to get the intersection points of a line and a curve which was fit to data?

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Hi,
I have a line and a curve that was fit to a data. I also get Coefficients of Equation of the Curve, but don't know how to find its equation to make two equations equal to find the points of the tangency. Could someone give me some recommends?
Here is my code:
clc
array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
0 150 300 394 450 540 600 631 696 745 750 865];
x=linspace(array(1,1),array(1,end),101)
y=interp1(array(1,:),array(2,:),x,'pchip')
x=transpose(x)
y=transpose(y)
%
f=fit(y,x,'pchip')
a=coeffvalues(f)
plot(f,y,x)
hold on
% Equation of line that pass through origin
x1=0:1000;
slope=tan(51.5*pi/180);
y1=slope*x1
plot(x1,y1)

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 24 Oct 2014
Edited: Andrei Bobrov on 25 Oct 2014
one way with Curve Fitting Toolbox
array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
0 150 300 394 450 540 600 631 696 745 750 865];
x = array([2 1],:)';
f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
xx = fzero(@(x)f(x)/x - df(x),[1 750]);
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,x1*df(xx),xx,f(xx),'ro');
well, more
f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
k = tand(10);
xx = fzero(@(x)df(x) - k,[1 x(end,1)]);
you line: y = k*x + b
b = f(xx) - k*xx;
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,k*x1 + b,xx,f(xx),'ro');
  8 Comments
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Hussein Qenawy on 13 Apr 2019
Edited: Hussein Qenawy on 13 Apr 2019
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