, 2012

b) Answer: $S = 27/4$.

We find the coordinates $(x_T; y_T)$ of the point of intersection of the parabola $y = a x^2$ and the hyperbola $y = 1/x$. We have $a x_T^2 = 1/x_T$, so $x_T = 1/\sqrt{a} = 1/t$ and $y_T = 1/x_T = t$. The equation of the tangent to the parabola at point $P(x_P; y_P)$ has the form $y - y_P = 2a x_P(x - x_P)$, and the equation of the tangent to the hyperbola at point $Q(x_Q; y_Q)$ has the form $y - y_Q = (-1/x_Q^2)(x - x_Q)$. Since the tangent is a common tangent for both the graphs, we have
$$2a x_P = -1/x_Q^2, \qquad (1)$$
$$-2a x_P^2 + y_P = 1/x_Q + y_Q. \qquad (2)$$
Since $y_P = a x_P^2$ and $y_Q = 1/x_Q$, the obtained equality has the form
$$-a x_P^2 = 2/x_Q. \qquad (2)$$
From equations (1), (2) it follows that $x_Q = -1/(2t)$, $x_P = -2/t$, and so $y_Q = -2t$, $y_P = 4t$. Therefore,
$$T(\frac{1}{t}; t), \quad P(-\frac{2}{t}; 4t), \quad Q(-\frac{1}{2t}; -2t). \qquad (3)$$
Let $M, K, L$ be the midpoints of the sides $PQ, PT, TQ$ respectively. Then
$$M(-\frac{5}{4t}; t), \quad K(-\frac{1}{2t}; \frac{5}{2}t); \quad L(\frac{1}{4t}; -\frac{t}{2}).$$
Comparing the obtained values with (3), we note that $MT$ parallel to the axis $Ox$, since $y_M = y_T$, and $QK$ parallel to $Oy$, since $y_Q = y_K$. Therefore, medians $MT$ and $QK$ in the triangle $PQT$ are perpendicular, as required.
Let $H$ be a centroid of the triangle $PQT$. Then
$$S(PQT) = 2S(QKT) = [TH \perp QK] = 2 \cdot \frac{1}{2} QK \cdot TH = \\ = [TH = \frac{2}{3}MT] = \frac{2}{3}QK \cdot TM.$$
Since $TM \parallel Ox$, we have $TM = 1/t - (-5/(4t)) = 9/(4t)$, Similarly, since $QK \parallel Oy$, we have $QK = 5t/2 - (-2t) = 9t/2$. Therefore,
$$S(PQT) = \frac{2}{3} \cdot \frac{9}{4t} \cdot \frac{9t}{2} = \frac{27}{4}.$$