How far does the secant of the parabola $y=x^{2}$ passing through the points $\left(n, n^{2}\right)$ and $\left(n+1,(n+1)^{2}\right)$ extend from the tangent to the parabola that is parallel to it? In what ratio does the projection of the point of tangency divide the chord formed on the secant?
The difference between the x-coordinates of the endpoints of the chord is 1, so the slope of the chord is equal to the difference of the y-coordinates, which is $2n+1$. Thus, the equation of the chord is

$$
y=(2 n+1)(x-n)+n^{2}=(2 n+1) x-\left(n^{2}+n\right)
$$

We know that the slope of the tangent line at any point on the parabola $y=x^{2}$ is twice the x-coordinate of the point of tangency. Therefore, the coordinates of the point of tangency of the tangent line parallel to the given chord are:

$$
T\left(n+\frac{1}{2},\left(n+\frac{1}{2}\right)^{2}\right)=T\left(n+\frac{1}{2}, n^{2}+n+\frac{1}{4}\right)
$$

The desired distance is given by the distance from $T$ to its projection $U$ on the line (1), which, using the known formula, is

$$
\begin{gathered}
d=T U=\frac{\left(n^{2}+n+1 / 4\right)-(2 n+1)(n+1 / 2)+\left(n^{2}+n\right)}{\sqrt{(2 n+1)^{2}+1}}= \\
=\frac{-1}{4 \sqrt{4 n^{2}+4 n+2}}
\end{gathered}
$$

the absolute value of the expression.

![](https://cdn.mathpix.com/cropped/2024_05_02_013b67e4882a82239e71g-1.jpg?height=453&width=788&top_left_y=928&top_left_x=663)

To determine the desired ratio $k$, it is sufficient to calculate one of the coordinates of $U$, since the ratio of the segments of a line segment is equal to the ratio of the corresponding segments of its projection, and thus

$$
\frac{A U}{U B}=k=\frac{x_{U}-x_{A}}{x_{B}-x_{U}}=\frac{y_{U}-y_{A}}{y_{B}-y_{U}}
$$

(where $A$ and $B$ are the endpoints of the segment).

The equation of the line $T U$ is

$$
y=-\frac{1}{2 n+1}\left(x-n-\frac{1}{2}\right)+n^{2}+n+\frac{1}{4}
$$

By eliminating $y$ from (1) and (3),

$$
x_{U}=\frac{4 n^{3}+6 n^{2}+7 n / 2+3 / 4}{4 n^{2}+4 n+2}=n+\frac{1}{2}-\frac{2 n+1}{4\left(4 n^{2}+4 n+2\right)}
$$

and thus, with $x_{A}=n, x_{B}=n+1$ and (2),

$$
\begin{aligned}
& k=\frac{\frac{1}{2}-\frac{2 n+1}{4\left(4 n^{2}+4 n+2\right)}}{\frac{1}{2}+\frac{2 n+1}{4\left(4 n^{2}+4 n+2\right)}}=\frac{8 n^{2}+6 n+3}{8 n^{2}+10 n+5}= \\
& =1-\frac{4 n+2}{8 n^{2}+10 n+5}=1-\frac{1}{2 n+\frac{3}{2}+\frac{1}{2 n+1}}
\end{aligned}
$$

As the absolute value of $n$ increases, the absolute value of the second term decreases arbitrarily, so the ratio deviates less and less from 1.

The denominator of the second term in the last but one form is positive for all $n$, and the sign of the numerator changes as $n$ passes through $-1 / 2$, so for $n < -1 / 2$, $k > 1$, for $n > -1 / 2$, $k < 1$, and for $n = -1 / 2$, $k = 1$. It is easy to show that for $n = -1 / 2$, $U$ falls on the focus of the parabola.