1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)$ is $2 \sqrt{3}$, and the distance between the points of intersection of the graphs $y=x^{2}-2$ and $y=f(x)+1$ is $\sqrt{60}$. Find the distance between the points of intersection of the graphs of the functions $y=x^{2}-1$ and $y=f(x)+1$.
Answer: $2 \sqrt{11}$.

Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}=a x+b$, and in the second case - from the equation $x^{2}-2=a x+b+1$.

Consider the first case in more detail. The equation has the form $x^{2}-a x-b=0$, from which $x_{1,2}=\frac{a \pm \sqrt{a^{2}+4 b}}{2}$, $\left|x_{2}-x_{1}\right|=\sqrt{a^{2}+4 b}$. Since the points of intersection of the graphs lie on a line with a slope of $a$, the distance between the points is $\sqrt{a^{2}+1}$ times greater than $\left|x_{2}-x_{1}\right|$. Therefore, the distance between the points is $\sqrt{\left(a^{2}+1\right)\left(a^{2}+4 b\right)}$. Similarly, we find that in the second case, the distance between the points is $\sqrt{\left(a^{2}+1\right)\left(a^{2}+4 b+12\right)}$. From the condition, we obtain the system of equations $\left\{\begin{array}{l}\left(a^{2}+1\right)\left(a^{2}+4 b\right)=12, \\ \left(a^{2}+1\right)\left(a^{2}+4 b+12\right)=60,\end{array}\right.$ solving which, we find that $a^{2}=3, b=0$.

Let's find the required distance. The abscissas of the points of intersection are determined by the equation $x^{2}-a x-b-2=0$, so $\left|x_{2}-x_{1}\right|=\sqrt{a^{2}+4 b+8}=\sqrt{11}$, and the distance between the points of intersection is $\left|x_{2}-x_{1}\right| \sqrt{a^{2}+1}=\sqrt{11} \cdot 2=2 \sqrt{11}$.