For what value of $\lambda$ does the following equation represent a pair of straight lines:

$$
\lambda x^{2}+4 x y+y^{2}-4 x-2 y-3=0
$$
Let's arrange the left side of the equation according to powers of \( y \):

\[
y^{2}+(4 x-2) y+\lambda x^{2}-4 x-3=0 \ldots
\]

This equation represents a pair of lines if \( y \) can be expressed as a first-degree function of \( x \). For this, it is necessary and sufficient that the discriminant be the square of a first-degree expression in \( x \). Already now,

\[
D=(4 x-2)^{2}-4\left(\lambda x^{2}-4 x-3\right)=(16-4 \lambda) x^{2}+16 \equiv(A x+B)^{2}
\]

From this it follows:

\[
B^{2}=16, \quad 2 A B=0 \quad \text { so } \quad A=0
\]

that is,

\[
16-4 \lambda=0 \quad \text { and thus } \quad \lambda=4
\]

Accordingly, from (1),

\[
y=\frac{-(4 x-2) \pm 4}{2}=\left\{\begin{array}{c}
-2 x+3 \\
-2 x-1
\end{array}\right.
\]

so \(\quad 4 x^{2}+4 x y+y^{2}-4 x-2 y-3 \equiv(y+2 x-3)(y+2 x+1)\),

which means our equation represents a pair of lines given by \(\quad y+2 x-3=0\) and \(y+2 x+1=0\).

The two lines are parallel to each other.

(Degenerate parabola!)

Boromissza Jenö and Lestál Lajos (Bencés g. VIII. o. Esztergom.)

Note. In the given case, when \(\lambda=4\), the discriminant of the expression

\[
\lambda x^{2}+4 x y+y^{2}
\]

also vanishes. However, the vanishing of this discriminant means that we are dealing with a parabola-like curve.