Problem 9.7. Through points $A(0 ; 14)$ and $B(0 ; 4)$, two parallel lines are drawn. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.

What is $\frac{A L-A K}{B N-B M}$?

![](https://cdn.mathpix.com/cropped/2024_05_06_899cf5197845501d962eg-31.jpg?height=525&width=519&top_left_y=488&top_left_x=467)
Answer: 3.5.

Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.

The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the roots of the equation $k x+14=\frac{1}{x}$, which can be rewritten as $k x^{2}+14 x-1=0$. Similarly, the abscissas of points $M$ and $N$ (denoted as $x_{M}$ and $x_{N}$, respectively) are the roots of the equation $k x+4=\frac{1}{x}$, that is, $k x^{2}+4 x-1=0$. By Vieta's theorem, $x_{K}+x_{L}=-\frac{14}{k}$ and $x_{M}+x_{N}=-\frac{4}{k}$.

Let $\angle(K L, O x)=\alpha$. Then

$$
\frac{A L-A K}{B N-B M}=\frac{\frac{\left|x_{L}\right|}{\cos \alpha}-\frac{\left|x_{K}\right|}{\cos \alpha}}{\frac{\left|x_{N}\right|}{\cos \alpha}-\frac{\left|x_{M}\right|}{\cos \alpha}}=\frac{-x_{L}-x_{K}}{-x_{N}-x_{M}}=\frac{\frac{14}{k}}{\frac{4}{k}}=\frac{14}{4}=3.5
$$