Mathematics Asked by a20 on December 29, 2021
I have measured the parameters for a hyperbola and an ellipse, let us call them
$$
begin{cases} a^2x^2 + b^2y^2 = 1 \ c^2x^2 + d^2y^2 = 1 end{cases}
$$
and I have errors associated with each parameter a, b, c and d ($pmdelta a$, $pmdelta b$, $pmdelta c$, $pmdelta d$). I want to find the intersection point(s) (due to symmetry there is in practice only one solution), between these two curves, that is, I want to solve the equation:
$$
AX = Y
$$
where I have defined $A=begin{pmatrix}a^2 & b^2 \ c^2 & d^2end{pmatrix}$, $X=begin{pmatrix}x^2 \ y^2end{pmatrix}$, $Y=begin{pmatrix}1 \ 1end{pmatrix}$.
Now, solving the equation is done by simply taking the inverse of A. What I am struggling with however is how to incorporate the errors, such that I derive the covariance matrix for the estimated intersection point $(hat{x},hat{y})$.
Let $a,b,c,d$ be the "correct" parameters and $a + delta a$, ..., $ d + delta d$ the measured parameters including errors. The resulting matrix is $A + delta A$ where to first order $$ delta A approx pmatrix{2 a; delta a & 2 b ; delta bcr 2 c; delta c & 2 d; delta d}$$ and you compute $X + delta X = (A + delta A)^{-1} Y$. Again to first order, assuming $A$ is invertible, $$(A + delta A)^{-1} approx A^{-1} + A^{-1} (delta A) A^{-1}$$ so $$delta X approx A^{-1} (delta A) A^{-1} Y $$
Answered by Robert Israel on December 29, 2021
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