Eccentricity of the conic $x^{2}-4xy-2y^{2}+10x+4y=0$?

Question

Q:

How do I find the eccentricity of the conic $x^{2}-4xy-2y^{2}+10x+4y=0$?

Doubt:

How do I solve this, because it is not like the normal hyperbola but also not like the $xy=c^2$ type.

Answer 1

How do I find the eccentricity of the conic \begin{align} x^2-4xy-2y^2+10x+4y&=0 \tag{1}\label{1} \end{align}

A general equation is of the form

\begin{align} Ax^2+Bxy+Cy^2+Dx+Ey+F&=0 \tag{2}\label{2} \end{align}

represents a hyperbola, if the its discriminant $\Delta=B^2-4AC>0$. In case of \eqref{1} we have

\begin{align} A&=1 ,\quad B=-4 ,\quad C=-2 ,\quad D=10 ,\quad E=4 ,\quad \text{and }\quad F=0 , \end{align}
hence

\begin{align} \Delta&=(-4)^2-4\cdot1\cdot(-2) =24>0 , \end{align} so, the equation \eqref{1} indeed represents a hyperbola which the eccentricity is given by

\begin{align} e&= \sqrt{\frac{2\sqrt{(A-C)^2+B^2}}{\eta(A+C)+\sqrt{(A-C)^2+B^2}}} \tag{3}\label{3} , \end{align} where $\eta=1$ if the determinant of the matrix \begin{align} M&= \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix} \tag{4}\label{4} \end{align} is negative, and $\eta=-1$ if the $\det(M)>0$.

For \eqref{1} we have

\begin{align} M&= \begin{bmatrix} \phantom{-}1 & -2 & 5 \\ -2 & -2 & 2 \\ \phantom{-}5 & \phantom{-}2 & 0 \end{bmatrix} \tag{5}\label{5} ,\\ \det{M}&=6>0 , \end{align} hence $\eta=-1$ and

\begin{align} e&=\sqrt{\tfrac53} \approx 1.29 . \end{align}