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

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Khanh
Khanh 2014-10-24
编辑： Hussein Qenawy 2019-4-13
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)

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Andrei Bobrov
Andrei Bobrov 2014-10-24
编辑：Andrei Bobrov 2014-10-25
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');
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Hussein Qenawy
Hussein Qenawy 2019-4-13
编辑：Hussein Qenawy 2019-4-13
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