For the family λ x^2-3xy+y^2+λ x+3=0 , determine λ so that…
For the family $\lambda x^2-3xy+y^2+\lambda x+3=0$, determine $\lambda$ so that the equation represents: $1^\circ$ ellipses (real or imaginary); $2^\circ$ hyperbolas; $3^\circ$ a parabola; $4^\circ$ a point; $5^\circ$ a pair of intersecting lines; $6^\circ$ a pair of parallel lines.
Explanation of Method
Classifying a one-parameter family of conics by the signs of δ and Δ.
Step 1: Invariants
\[ \delta=\lambda-\tfrac94,\qquad \Delta=\begin{vmatrix}\lambda&-\tfrac32&\tfrac\lambda2\\-\tfrac32&1&0\\\tfrac\lambda2&0&3\end{vmatrix}=-\frac{\lambda^2}{4}+3\lambda-\frac{27}4=-\frac{(\lambda-3)(\lambda-9)}{4}. \]
Step 2: Classification
| Type | Condition | $\lambda$ |
|---|---|---|
| real ellipses | $\delta\gt 0,\ A\Delta\lt 0$ | $\tfrac94\lt \lambda\lt 3$ or $\lambda\gt 9$ |
| imaginary ellipses | $\delta\gt 0,\ A\Delta\gt 0$ | $3\lt \lambda\lt 9$ |
| hyperbolas | $\delta\lt 0$ ($\Delta\ne0$ automatically) | $\lambda\lt \tfrac94$ |
| parabola | $\delta=0,\ \Delta=-\tfrac{81}{64}\ne0$ | $\lambda=\tfrac94$ |
| point | $\delta\gt 0,\ \Delta=0$ | $\lambda=3$ or $\lambda=9$ |
| intersecting lines | $\delta\lt 0,\ \Delta=0$ | none |
| parallel lines | $\delta=0,\ \Delta=0$ | none |
Since $A=\lambda\gt 0$ on the whole ellipse range, the sign test $A\Delta\lt 0$ reduces to $\Delta\lt 0$.
Step 3: The two points
$\lambda=3$: the centre equations $6x-3y+3=0$ and $-3x+2y=0$ give $(-2,-3)$, and $F(-2,-3)=12-18+9-6+3=0$. Since $\delta\gt 0$, the quadratic form is definite. The conic is therefore the single real point $(-2,-3)$: two complex-conjugate lines meet there.
$\lambda=9$: in the same way the point is $\left(-\tfrac23,-1\right)$, with $F=4-2+1-6+3=0$.
\[\lambda=3:\ \{(-2,-3)\},\qquad \lambda=9:\ \left\{\left(-\tfrac23,-1\right)\right\}\]
Result
\[\boxed{\,\lambda=3:\ \{(-2,-3)\},\qquad \lambda=9:\ \left\{\left(-\tfrac23,-1\right)\right\}\,}\]
Remark
Clarification: the source's wording in $4^\circ$ (“point or two real lines”, and the condition $\Delta/A\lt 0$) is imprecise. With $\delta\gt 0$ and $\Delta=0$ the curve is always exactly one real point, never two real lines. For $5^\circ$ and $6^\circ$ there is no solution: $\Delta=0$ forces $\lambda\in\{3,9\}$, where $\delta\gt 0$, and $\delta=0$ forces $\lambda=\tfrac94$, where $\Delta\ne0$.