I have represented the conic $C$ in the Euclidean plane described by the equation $$3x^{2}+ 2y^{2}+ 7xy + 4x + 5y + 3 = 0$$ as a symmetric matrix. \begin{pmatrix} 3 & 7/2 & 2\\ 7/2 & 2 & 5/2\\ 2 & 5/2 & 3\end{pmatrix}.
Now I am asked to show that $C$ is nonsingular and identify the type of conic. I think I have shown it is nonsingular using the fact that $A$ is singular, i.e the determinant of $A$ is $0$ if and only if the polynomial $X^{t}AX$ can be factorized over the complex numbers into two linear factors. I did this by calculating $X^{t}AX$ to obtain $17/2x^{2}+8y^{2} +15z^{2}$ and stated that this cannot be factorized thus it is non singular.
I need help with identifying the type of conic. I initially took the result from above, $17/2x^{2}+8y^{2} +15z^{2}$ and converted it back into Cartesian but could not get it in the form of any conic. Any tips?
Let the matrix you have described be called by $A_Q$ and $Q$ be the type of conic.
If $\det A_Q =0$, then the conic is degenerate. If $\det A_Q\neq 0$, so that the conic is not degenerate, we can see what type of conic it is by computing the minor, $\det A_{33}$. Thus,
$(1)$ It is a hyperbola iff $\det A_{33}<0$.
$(2)$ It is a parabola iff $\det A_{33} =0$.
$(3)$ It is an ellipse iff $\det A_{33} >0$.
In your case, we have $\det A_{33} = 3\times 2-(\frac{7}{2})^{2} < 0$. So it is a hyperbola. Hope it helps.