Parametric radius of a circle

Hey everyone!
Ive been trying to solve "the envelope problem" i uploaded a few days ago and ive made huge progress. However im now stuck on some silly little thing. I need to find the parametric equation for circles radius (later many circles will be drawn with that).
My code so far is :
syms a b p x y L
F = [p/2,0];
xpara = [x sqrt(2*p*x)];
eqn1 = a*F(1,1)+b == 0;
eqn2 = a*x+b == xpara(1,2);
eqns = [eqn1 eqn2];
%L - lambda.
eqns = [a*F(1,1)+b == 0, subs(a*x+b,x,L) == subs(xpara(1,2),x,L)];
S=solve(eqns,[a b],'ReturnConditions',true);
S.a;
S.b;
sirge =simplify(S.a*x+S.b);
loikepunktx = solve(sirge==-xpara(1,2),x,'ReturnConditions',true);
%sirge is a line equation
loikepunktx.x=loikepunktx.x(2,1);
loikepunktx.x;
sirge1 = simplify(S.a * loikepunktx.x + S.b);
sirge1;
LP = [loikepunktx.x, sirge1];
%Center = KP
KP = subs(simplify((xpara + LP)/2),x,L);
%ok so far
xpara;
KP;
In wolframalpha the solution would be
pR = Simplify[(xpara - KP) . (xpara - KP)] /. x -> \[Lambda]
and it gives the answer :
In MatLab however, im always getting a dot (i.e (x,y)) not the formula. Matlab doesnt simplify it like wolfram but the answer should be similar/ simplifiable to that form.
Probably going to be asking a few more questions while trying to progress but i am close to the end :)
Edit: i just did it the easy way:
R = subs(sqrt((xpara(1,1)-KP(1,1))^2+(xpara(1,2)-KP(1,2))^2),x,L)
But it doesnt give the same answers as in wolfram. The wolframalpha answer is squared of what matlab gives me. I dont know whether my WA has a mistake in it or my matlab code.

3 comentarios

darova
darova el 27 de Mayo de 2021
MATLAB has weak symbolic toolbox. WHat about numerical approach?
Stefan Juhanson
Stefan Juhanson el 28 de Mayo de 2021
Can not be approached numerically since in the end i need to plot a group of circle shaped envelopes. Theres a parabola y^2 = 2*p*x. And on its focalchords are the diameters of the sought circles. The task is to get the equation for the circles envelope group and plot it. The end plot looks like this:
But this was done in wolframalpha so now i am trying to convert the code into MatLab
Stefan Juhanson
Stefan Juhanson el 28 de Mayo de 2021
I found the reason why i get it squared in wolfram alpha. I used scalar multiplication in WA and if a vector is multiplied by itself it gives R^2 not R.

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el 26 de Mayo de 2021

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