Projection of a point onto a closed convex

I have a point P(50, 5000) and a curve (x.^2 - 2*x - 1) with x = [0:100]. How do I find the closest point on that curve to point P?

Thank you.

Answer 1

[Bruno Luong]

Use this FEX

Q=[50;5000];

% Projection candidates on x.^2-2*x-y-1 = 0
P=ConicPrj(Q,[1 0; 0 0],[-2;-1],-1);
% Find the closest
[~,loc]=min(sum((P-Q).^2,1));
P=P(:,loc);

x=P(1)
y=P(2)

The projection is

x = 71.723733062992125

y = 4.999846418365362e+03

[Donigo Fernando Sinaga]

Error using -

Matrix dimensions must agree.

Thank you, but I got this error, how can I fix this? I think P can't be subtracted by Q.

[Bruno Luong]

Runing on older version of MATLAB, you must replace

P-Q

by

bsxfun(@minus,P,Q)

[Donigo Fernando Sinaga]

How to change my function to that ConicPrj function input? What is a, b, and c?

[Bruno Luong]

as written in the H1 line of ConicPrj , the conic is implicit equation -I change x to v here to avoid confusion with your x

E = { v such that: v'*A*v + b'*v + c = 0}

In your case, if I define v := [x;y];

the equation of parabolic is

x^2- 2*x - y -1 = 0

Meaning

x*(1)*x + x*0*y + y*0*x + y*0*y + (-2)*x + (-1)*y -1 = 0
   ^        ^       ^       ^       ^        ^     ^
A(1,1)   A(1,2)  A(2,1)  A(2,2)    b(1)     b(2)   c

or equivalently

v' * [1 0; 0 0] * v + [-2; -1]'*v - 1 = 0;

Therefore

 A=[1 0; 0 0]; b  = [-2; -1]; and c = -1;