Problem 1104

I. Solution. Let's assume that the answer to the problem is affirmative, and the point $P\left(t, t^{2}\right)$ on the parabola $p$ and the point $Q\left(u, 2 u^{2}-7 u / 2+57 / 16\right)$ on the parabola $q$ satisfy the condition. Then the tangent to $p$ at $P$ and the tangent to $q$ at $Q$ are parallel to each other.

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Since the ordinate is a differentiable function of the abscissa for both curves, the tangent's slope is given by the derivative at the respective point, which is $2t$ for $p$ and $P$, and $4u - 7/2$ for $q$ and $Q$, and these two are equal:

$$2t = 4u - 7/2, \quad t = 2u - 7/4$$

On the other hand, the line $PQ$ is perpendicular to both tangents, i.e., $PQ$ is a common normal to both parabolas. The slope of the normal to $q$ at $Q$ is $-1/(4u - 7/2)$, provided the denominator is non-zero, i.e., $u \neq 7/8$. Thus, the equation of the normal is:

$$y - \left(2u^2 - \frac{7}{2}u + \frac{57}{16}\right) = \frac{-1}{4u - 7/2}(x - u)$$

and this equation is satisfied by the coordinates of $P$. Substituting the coordinates of $P$ in terms of $u$ using (1), we get an equation for $u$:

$$\begin{gathered} \left(2u - \frac{7}{4}\right)^2 - 2u^2 + \frac{7}{2}u - \frac{57}{16} = \frac{-1}{4u - 7/2}(u - 7/4) \\ 8u^3 - 21u^2 + \frac{45}{4}u = 8u\left(u^2 - \frac{21}{8}u + \frac{45}{32}\right) = 8u\left(u - \frac{15}{8}\right)\left(u - \frac{3}{4}\right) = 0 \end{gathered}$$

Thus, the appropriate values of $u$, and then $t$ using (1), and the corresponding pairs of points $P, Q$ that satisfy the condition are as follows (arranged in columns):

$$\begin{array}{lll} u_1 = 0 & u_2 = \frac{3}{4} & u_3 = \frac{15}{8} \\ t_1 = -\frac{7}{4} & t_2 = -\frac{1}{4} & t_3 = 2 \\ P_1\left(-\frac{7}{4} ; \frac{49}{4}\right) & P_2\left(-\frac{1}{4} ; \frac{1}{16}\right) & P_3(2 ; 4) \\ Q_1\left(0 ; \frac{57}{16}\right) & Q_2\left(\frac{3}{4} ; \frac{33}{16}\right) & Q_3\left(\frac{15}{9} ; \frac{129}{32}\right) \end{array}$$

Remark. The precise determination of the pairs of points $P_i, Q_i (i=1,2,3)$ is based on recognizing that $u=0$ is a root of the cubic equation (3). If we had eliminated $u$ instead of $t$, this fortunate situation would not have arisen. It is therefore useful to sometimes try such approaches.

II. Solution (outline). Any two parabolas, $v_1$ and $v_2$, are similar to each other (in shape), as they are determined by a single linear parameter, $p_1$ and $p_2$, respectively. By scaling $v_1$ by a factor of $\lambda = p_2 / p_1$ and appropriately translating it, we obtain $v_2$. In this transformation, the foci $F_1$ and $F_2$, and the vertices $C_1$ and $C_2$, correspond to each other in pairs. Furthermore, the image of any point $A_1$ on $v_1$ in $v_2$ is the point $A_2$ where the ray from $F_2$ that forms the same angle and direction as the angle $C_1 F_1 A_1$ intersects $v_2$.

If the axes of the two parabolas are parallel (and lie in the same plane), the tangent at any point $A_1$ on $v_1$ is parallel to the tangent at the corresponding point $A_2$ on $v_2$. In this case, $v_1$ and $v_2$ are in a central similarity position relative to each other—unless $C_1 F_1 F_2 C_2$ forms a parallelogram—the center of similarity is the intersection point $H$ of the lines $F_1 F_2$ and $C_1 C_2$. (In the special case, $v_1$ and $v_2$ can be translated into each other).

In our case, we are looking for a pair of points $P, Q$ on the parabolas $p$ and $q$ with parallel axes such that the tangents at these points are parallel. Therefore, $P$ and $Q$ are corresponding points in the similarity, as a parabola can have exactly one tangent parallel to any given direction. Thus, the line $PQ$ passes through $H$. Based on this, our plan is as follows: we calculate the coordinates of $H$, write the equation of the normal $n$ to $q$ at the point $Q$ with abscissa $u$, and then we get an equation for $u$ by substituting the coordinates of $H$ into the equation of $n$.

Now, the given parabola $p$ has its vertex $C_1$ at the origin, and its parameter is $p_1 = 1/2$. Similarly, by appropriately rearranging the equation of $q$:

$$$$
\left(x - \frac{7}{8}\right)^2 = \frac{1}{2