Consider all segments PQ of length 2 such that the end points P and Q are on the parabola $y = x ^ { 2 }$. Denote the mid-point of the segment PQ by M. Among all M, we are to find the coordinates of the ones nearest to the $x$-axis.
Let us denote the coordinates of the end points of segment PQ by $\mathrm{ P }\left( p , p ^ { 2 } \right)$ and $\mathrm{ Q }\left( q , q ^ { 2 } \right)$. Then the $y$-coordinate $m$ of M is
$$m = \frac { p ^ { 2 } + q ^ { 2 } } { \mathbf { M } } .$$
Next, since $\mathrm{ PQ } = 2$, then
$$( p - q ) ^ { 2 } + \left( p ^ { 2 } - q ^ { 2 } \right) ^ { 2 } = \mathbf { N }$$
by the Pythagorean theorem.
Now, when we set $t = p q$, we obtain from (1) and (2) the quadratic equation in $m$
$$\mathbf { O } m ^ { 2 } + m - \mathbf { P } t ^ { 2 } - t - \mathbf { Q } = 0 .$$
When we solve this for $m$, noting that $m > 0$, we have
$$m = - \frac { 1 } { \mathbf { R } } + \sqrt { \left( t + \frac { 1 } { \mathbf { S } } \right) ^ { 2 } + \mathbf{T} } .$$
This shows that $m$ is minimized when $t = - \dfrac { 1 } { \mathbf{U} }$. In this case, $p q = - \dfrac { 1 } { \mathbf{U} }$ and $p ^ { 2 } + q ^ { 2 } = \dfrac { \mathbf { V } } { \mathbf { V } }$, and so we have $p + q = \pm \mathbf { W }$.
Thus the coordinates of the M nearest to the $x$-axis are $\left( \pm \dfrac { 1 } { \mathbf { X } } , \dfrac { \mathbf { Y } } { \mathbf { Z } } \right)$.