In x^2+λ xy+y^2-4x-4y+1=0 determine λ so that the equation…
In $x^2+\lambda xy+y^2-4x-4y+1=0$ determine $\lambda$ so that the equation represents: $1^\circ$ a pair of lines; $2^\circ$ a parabola; $3^\circ$ a rectangular hyperbola.
Explanation of Method
Choosing λ so that the conic is a pair of lines, a parabola, or a rectangular hyperbola.
Step 1: Invariants
\[ \delta=1-\frac{\lambda^2}{4},\qquad \Delta=\begin{vmatrix}1&\tfrac\lambda2&-2\\\tfrac\lambda2&1&-2\\-2&-2&1\end{vmatrix}=-\frac{\lambda^2}{4}+4\lambda-7=-\frac{(\lambda-2)(\lambda-14)}{4}. \]
Step 2: $1^\circ$ Pair of lines ($\Delta=0$)
$\lambda=2$ ($\delta=0$): with $t=x+y$, $t^2-4t+1=0$, so two parallel lines:
\[x+y=2+\sqrt3,\qquad x+y=2-\sqrt3\]
$\lambda=14$ ($\delta=-48\lt 0$): two intersecting lines. The slopes come from $t^2-14t+1=0$, $t=7\pm4\sqrt3$:
\[\big(x+(7+4\sqrt3)y-2-\sqrt3\big)\big(x+(7-4\sqrt3)y-2+\sqrt3\big)=0\]
The lines meet at the centre $\left(\tfrac14,\tfrac14\right)$, which solves $2x+14y-4=0$ and $14x+2y-4=0$.
Step 3: $2^\circ$ Parabola ($\delta=0,\ \Delta\ne0$)
$\delta=0$ gives $\lambda=\pm2$. $\lambda=2$ is excluded because $\Delta=0$ there. For $\lambda=-2$, $\Delta=-16\ne0$.
\[\lambda=-2:\quad (x-y)^2=4(x+y)-1\]
In the rotated axes $u=\tfrac{x+y}{\sqrt2}$, $v=\tfrac{y-x}{\sqrt2}$ this becomes $v^2=2\sqrt2\,u-\tfrac12$. The axis is $y=x$ and the vertex is $\left(\tfrac18,\tfrac18\right)$.
Step 4: $3^\circ$ Rectangular hyperbola
A hyperbola is rectangular iff its asymptotes are perpendicular, iff the eigenvalues of the quadratic form are opposite, iff $A+C=0$. The trace $A+C$ is invariant under rotation. Here $A+C=2$ for every $\lambda$ (the eigenvalues are $1\pm\tfrac{\lambda}{2}$), so:
\[\text{no value of }\lambda\text{ gives a rectangular hyperbola}\]
Step 5: Complete classification
| $\lambda$ | Curve |
|---|---|
| $|\lambda|\lt 2$ | real ellipse ($\Delta\lt 0$, $A=1\gt 0$) |
| $\lambda=-2$ | parabola |
| $\lambda=2$ | two parallel lines |
| $\lambda\lt -2$, $2\lt \lambda\lt 14$, $\lambda\gt 14$ | hyperbola |
| $\lambda=14$ | two intersecting lines |
Result
\[\boxed{\,x+y=2+\sqrt3,\qquad x+y=2-\sqrt3\,}\]
\[\boxed{\,\big(x+(7+4\sqrt3)y-2-\sqrt3\big)\big(x+(7-4\sqrt3)y-2+\sqrt3\big)=0\,}\]
\[\boxed{\,\lambda=-2:\quad (x-y)^2=4(x+y)-1\,}\]
\[\boxed{\,\text{no value of }\lambda\text{ gives a rectangular hyperbola}\,}\]
Remark
Corrections:
- The source's summary lists hyperbolas only for “$\lambda\lt -2$ or $2\lt \lambda\lt 14$”, which drops $\lambda\gt 14$. There $\delta\lt 0$ and $\Delta\ne0$, so these are hyperbolas too.
- The unfinished formula “$\tan^2\varphi=-\delta/?$” in the source is unnecessary. The criterion $A+C=0$ settles $3^\circ$.