Consider the following system of equations:

$$
\begin{aligned}
& y = mx \pm a \ldots \\
& \frac{1}{x} - \frac{1}{y} = \frac{1}{a} \cdots
\end{aligned}
$$

Here, $a$ represents a given positive number, and $m$ represents an arbitrary real number.

$1^{0}$. Solve the system of equations and determine for which values of $m$ there are real solutions.

$2^{0}$. Plot the curves corresponding to (1) and (2), and use this to verify the result of the analysis in $1^{0}$.
$1^{0}$. Essentially, we have two systems of equations, depending on whether we take $a$ with a + or - sign.
I.

$$
y=m x+a \ldots
$$

$$
a(y-x)=x y \ldots
$$

Eliminating $y$ results in:

$$
m x^{2}+a(2-m) x-a^{2}=0 \ldots
$$

The roots of 3) are real if

$$
a^{2}(2-m)^{2}+4 a^{2} m \geqq 0, \quad \text { i.e., } \quad a^{2}\left(4+m^{2}\right)>0
$$

This condition is always satisfied; the system of equations I has two real solutions for any value of $m$.

II.

$$
\begin{gathered}
y=m x-a \ldots \\
a(y-x)=x y \ldots
\end{gathered}
$$

Eliminating $y$ results in: $m x^{2}-\max +a^{2}=0 \ldots$

The roots of 4) are real if

$$
m^{2} a^{2}-4 m a^{2} \geqq 0 \quad \text { i.e., } \quad m(m-4) \geqq 0
$$

In this case, the roots are real if $m \leqq 0$ or $m \geqq 4$.

$2^{0}$. From 2), $y=\frac{a x}{a-x}=\frac{a(x-a)+a^{2}}{a-x}=-a+\frac{a^{2}}{a-x}$.

If $x$ approaches infinity, the value of the function approaches $-a$, and the curve approaches the line $y=-a$ (and touches it at infinity if $x= \pm \infty$).

The value of $y$ approaches $+\infty$ if $x$ approaches $a$ from the side of values less than $a$, but approaches $-\infty$ if $x$ approaches $a$ from the side of values greater than $a$. The curve touches the line $x=a$ at infinity (if $y= \pm \infty$).

![](https://cdn.mathpix.com/cropped/2024_05_02_9dce1d50e628f33191c1g-1.jpg?height=561&width=512&top_left_y=1487&top_left_x=796)

The curve has two perpendicular asymptotes, hence it is a hyperbola with equal sides.

The line $y=m x+a$ passes through the point $K(x=0, y=+a)$; its slope is $m$. Any line passing through point $K$ intersects the curve at two points. If $m=0$, one intersection point is $P\left(x=\frac{a}{2}, y=a\right)$, and the other is at infinity.

If $m=\infty$, one intersection point is the origin $(x=0, y=0)$, and the other is at infinity.

The line $y=m x-a$ passes through the point $L(x=0, y=-a)$. Among the lines passing through point $L$, the one with $m=0$ (parallel to the $X$-axis) is an asymptote of the curve. The one with $m=4$ is tangent to the curve. Indeed, in this case, from 4),

$$
4 x^{2}-4 a x+a^{2}=0 \quad \text { i.e., } \quad(2 x-a)^{2}=0
$$

$x=\frac{a}{2}$ is a double root of this equation; thus, the line $y=4 x-a$ is tangent to the curve at the point $P\left(x=\frac{a}{2}, y=a\right)$.

Among the lines passing through point $L$, those that lie within the acute angle formed by the asymptote $y=-a$ ($X^{\prime} L X$) and the tangent $y=4 x-a$ do not intersect the curve. (For these, $0<m<4$).