For what value of $\lambda$ does the equation

$$
\lambda x^{2}+4 x y+y^{2}-4 x-2 y-3=0
$$

represent a pair of lines?
I. Solution: The given equation will represent a pair of lines if, when solving for $y$ as a function of $x$, we obtain two linear equations in $y$. Therefore, let's rearrange the equation in terms of powers of $y$ and solve for $y$:

$$
y^{2}+2 y(2 x-1)+x(\lambda x-4)-3=0
$$

from which

$$
\begin{gathered}
y=\frac{-2(2 x-1) \pm \sqrt{4(2 x-1)^{2}-4 x(\lambda x-4)+12}}{2} \\
=1-2 x \pm \sqrt{(4-\lambda) x^{2}+4}
\end{gathered}
$$

The obtained equation will only be linear in $x$ if the coefficient of $x^{2}$ under the square root is 0 (since the two-term expression under the square root cannot be expressed as a perfect square).

Thus, the given equation represents a pair of lines for $\lambda=4$. From the expression obtained for $y$, the equations of the two lines are

$$
y=-2 x+3 \quad \text { and } \quad y=-2 x-1
$$

Katona Gyula (Bp. VIII., Kandó híradásip. t. III. o. t.)

II. Solution: The equation of a "pair of lines" (i.e., a degenerate conic section) can be derived by multiplying the equations of the two lines. The equation of neither of the lines in the pair can be of the form $x=a$ (constant), because multiplying such an equation by another linear equation in $x$ and $y$ would not yield a $y^{2}$ term. Therefore, the equations of the two lines can be written as $y=m_{1} x+b_{1}$ and $y=m_{2} x+b_{2}$. The equation of the degenerate conic section formed by these two lines (by multiplying their zero-referenced forms):

$$
\begin{gathered}
\left(y-m_{1} x-b_{1}\right)\left(y-m_{2} x-b_{2}\right)= \\
=m_{1} m_{2} x^{2}-\left(m_{1}+m_{2}\right) x y+y^{2}+\left(m_{1} b_{2}+m_{2} b_{1}\right) x-\left(b_{1}+b_{2}\right) y+b_{1} b_{2}=0
\end{gathered}
$$

By comparing the coefficients of the given equation and the one just obtained, we arrive at the following equations:

$$
\begin{aligned}
m_{1} \cdot m_{2} & =\lambda, & -\left(b_{1}+b_{2}\right)=-2, \\
-\left(m_{1}+m_{2}\right) & =4, & b_{1} b_{2}=-3, \\
m_{1} b_{2}+m_{2} b_{1} & =-4 . &
\end{aligned}
$$

The unknowns can be calculated from the obtained system of equations. From the last two equations, $b_{1}$ and $b_{2}$ are the two roots of the equation $x^{2}-2 x-3=0$: $b_{1}=3, b_{2}=-1$. Using these, from the second and third equations, we get $m_{1}=m_{2}=-2$. From this, $\lambda=m_{1} m_{2}=4$.

Thus, the given equation will represent a pair of lines for $\lambda=4$. Knowing $m_{1}, m_{2}, b_{1}, b_{2}$, we can verify that the solution is correct by multiplying the equations of the lines $[(y+2 x-3) \cdot(y+2 x+1)=0]$.