Anisotropic, two-dimensional Gaussians, are described by the equation (from here):
$f(x,y)=A\exp \left( -(a(x-x_0)^2 + 2b(x-x_0)(y-y_0) + c(y-y_0)^2 )\right)$
where A is the amplitude of the peak, $x_0$ and $y_0$ are the $x$ and $y$ coordinates of the peak, respectively, and:
$a=\frac{\cos^2(\theta)}{2\sigma_{\rm maj}^2} + \frac{\sin^2(\theta)}{2\sigma_{\rm min}^2}$, $b=-\frac{\sin(2\theta)}{4\sigma_{\rm maj}^2} + \frac{\sin(2\theta)}{4\sigma_{\rm min}^2}$, $c=\frac{\sin^2(\theta)}{2\sigma_{\rm maj}^2} + \frac{\cos^2(\theta)}{2\sigma_{\rm min}^2}$
where $\theta$ is the angle of clockwise rotation for the major axis from the $x$-axis, $\sigma_{\rm maj}$ is the standard deviation of the major axis and $\sigma_{\rm min}$ is the standard deviation of the minor axis.
Given two, 2-dimensional Gaussians, $F(x,y)$ and $G(x,y)$, their convolution yields another 2-dimensional Gaussian distribution, $H(x,y)$, i.e. $F*G=H$.
In the context of this question, the properties of G and H are both known, whereby they are both anisotropic, and $A_H \neq A_G, \theta_H \neq \theta_G, \sigma_{\rm maj,H} \neq \sigma_{\rm maj,G}, \sigma_{\rm min,H} \neq \sigma_{\rm min,G}$)
My question is this, how do I calculate $\theta_F, \sigma_{\rm maj,F}$, and $\sigma_{\rm min,F}$ ?
Note: In the case where $G(x,y)$ is isotropic (i.e. 'circular'), the solution is simply $\theta_{\rm F}=\theta_{\rm H}$, $\sigma_{\rm maj,F} = \sqrt{\sigma_{\rm maj,H}^2 - \sigma_{\rm G}^2}$, and $\sigma_{\rm min,F} = \sqrt{\sigma_{\rm min,H}^2 - \sigma_{\rm G}^2}$
Hint: $$ \eqalign{ & a\left( {x - x_0 } \right)^2 + 2b\left( {x - x_0 } \right)\left( {y - y_0 } \right) + c\left( {y - y_0 } \right)^2 = \cr & = \xi ^2 + {{2b} \over {\sqrt {ac} }}\xi \eta + \eta ^2 = \cr & = \xi ^2 + {{2b} \over {\sqrt {ac} }}\xi \eta + {{b^2 } \over {ac}}\eta ^2 + \left( {1 - {{b^2 } \over {ac}}} \right)\eta ^2 = \cr & = \left( {\xi + {b \over {\sqrt {ac} }}\eta } \right)^2 + \left( {1 - {{b^2 } \over {ac}}} \right)\eta ^2 \cr} $$
You may want to apply matrix notation to the above ..
p.s.: do not forget the minus sugn in the $\exp$