Let's determine the distance between the tangents parallel to the line $y=x$ for the parabolas given by the equations $y=x^{2}+1$ and $x=y^{2}+1$.

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Let's determine the distance between the tangents parallel to the line $y=x$ for the parabolas given by the equations $y=x^{2}+1$ and $x=y^{2}+1$.
The two parabolas are symmetrical with respect to the line $y=x$. The tangent $e$ to the parabola $y=x^{2}+1$ that is parallel to the line $y=x$ has a slope of 1, and its equation is $y=x+b$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7dc644ef82793557bbdeg-1.jpg?height=493&width=539&top_left_y=233&top_left_x=782)

At the point of tangency $E$, the values of the two functions are equal:

\[
\begin{aligned}
& x^{2}+1=x+b, \quad \text { hence } \\
& x^{2}-x+(1-b)=0
\end{aligned}
\]

The quadratic equation has one root if its discriminant is 0: \(D=1-4(1-b)\), hence \(b=\frac{3}{4}\), so the equation of the tangent is \(y=x+\frac{3}{4}\).

Let's calculate the coordinates of the point of tangency $E$: \(x^{2}+1=x+\frac{3}{4}, x^{2}-x+\frac{1}{4}=0 ; x=\frac{1}{2}, y=\frac{1}{2}+\frac{3}{4}=\frac{5}{4}\).

It is sufficient to determine the distance of the point \(E\left(\frac{1}{2}, \frac{5}{4}\right)\) from the line \(y=x\), which is twice the distance between the parallel tangents. Let's set up a line perpendicular to \(y=x\) through the point \(E\). Its equation is: \(y-\frac{5}{4}=-\left(x-\frac{1}{2}\right)\), rearranging: \(y=-x+\frac{7}{4}\), the coordinates of the intersection point \(M\) with the line \(y=x\) are: \(M\left(\frac{7}{8}, \frac{7}{8}\right)\). The distance between points \(E\) and \(M\):

\[
\sqrt{\left(\frac{7}{8}-\frac{1}{2}\right)^{2}+\left(\frac{7}{8}-\frac{5}{4}\right)^{2}}=\frac{1}{8} \sqrt{18}=\frac{3}{8} \sqrt{2}
\]

The distance between the two tangents is \(d=\frac{3 \sqrt{2}}{4}\).

Anna Feher (Budapest, St. Stephen's Gymnasium, 9th grade)