Determine what curves are represented by x^2-2xy+λ…

Determine what curves are represented by $x^2-2xy+\lambda y^2-4x-6y+3=0$ for the various values of $\lambda$.

Step 1: Invariants

\[ \delta=\lambda-1,\qquad \Delta=\begin{vmatrix}1&-1&-2\\-1&\lambda&-3\\-2&-3&3\end{vmatrix}=(3\lambda-9)-9-2(3+2\lambda)=-\lambda-24 . \]

Step 2: Classification

\[\begin{cases}\lambda\gt 1: & \text{real ellipse}\\ \lambda=1: & \text{parabola}\\ -24\lt \lambda\lt 1\ \text{or}\ \lambda\lt -24: & \text{hyperbola}\\ \lambda=-24: & \text{two intersecting lines}\end{cases}\]

Step 3: Details

$\lambda=-24$:

\[ (x-6y-3)(x+4y-1)=0,\qquad\text{intersection }\left(\tfrac95,-\tfrac15\right). \]

$\lambda=1$: $\Delta=-25\ne0$. The quadratic part is $(x-y)^2$, so the axis is parallel to $(1,1)$. Completing the square:

\[ (x-y)^2-4x-6y+3=0\ \Longleftrightarrow\ (x-y-2)^2=10y+1 . \]

$\lambda\gt 1$: the centre is $\left(\tfrac{2\lambda+3}{\lambda-1},\tfrac5{\lambda-1}\right)$, and the value there is $F_0=-\tfrac{\lambda+24}{\lambda-1}\lt 0$. With a positive definite form this gives a real ellipse for every $\lambda\gt 1$. For example, $\lambda=2$ gives centre $(7,5)$ and $F_0=-26$.

Hyperbolas: the same formula for $F_0$ holds for $\lambda\lt 1$. $F_0$ changes sign at $\lambda=-24$: for $-24\lt \lambda\lt 1$, $F_0\gt 0$, and for $\lambda\lt -24$, $F_0\lt 0$. The two hyperbola families therefore lie in opposite pairs of angles between the asymptotes, and the degenerate line pair at $\lambda=-24$ separates them.