I have to prove that an affine trasformation $H_a=\begin{bmatrix} A & t \\ 0^T & 1\end{bmatrix}$ transforms a circle $C=\begin{bmatrix}1 & 0 & \frac{d}{2} \\ 0 & 1 & \frac{e}{2} \\ \frac{d}{2} & \frac{e}{2} & f\end{bmatrix}$ into an ellipse (so the matrix $C'=H_a^{-T}CH_a^{-1}$ represents an ellipse) and it cannot map the ellipse into a hyperbola. Is there a short way to prove it without computing all the products and the inverse matrices?
Thank you in advance
If $H_a$ is nonsingular, then so is $A$ and $$H_a^{-1}=\begin{bmatrix}A^{-1}&-A^{-1}\mathbf t\\0&1\end{bmatrix}.$$ Writing $C$ in block form and transforming it produces $$\begin{bmatrix}(A^{-1})^T&0\\-(A^{-1}\mathbf t)^T&1\end{bmatrix}\begin{bmatrix}I_2&\mathbf b\\\mathbf b^T&f\end{bmatrix}\begin{bmatrix}A^{-1}&-A^{-1}\mathbf t\\0&1\end{bmatrix}=\begin{bmatrix}(A^{-1})^TA^{-1}&\cdots\\\vdots&\ddots\end{bmatrix}.$$ $\det (A^{-1})^TA^{-1}=\frac1{(\det A)^2}\gt0$, therefore the transformed equation represents an ellipse.