69th Belarusian Mathematical Olympiad

a) First we find the values of $a$ for which the parabola is tangent to the hyperbola (see fig.). Let $\alpha$ be the abscissa of the tangency point. At this point the derivatives of functions $x^2-a$ and $1/x$ are equal, i.e. $2\alpha =-1/{\alpha }^2$, whence $\alpha =-\frac{1}{\sqrt[3]{2}}$ and the ordinate of the tangency point equals $\frac{1}{\alpha }=-\sqrt[3]{2}$. Then the parabola equation for this point gives $-\sqrt[3]{2}={\left(-\frac{1}{\sqrt[3]{2}}\right)}^2-a$, whence $a=\frac{3}{2}\sqrt[3]{2}$. It is clear that for $a<\frac{3}{2}\sqrt[3]{2}$ the parabola and the hyperbola have exactly one common point and for $a>\frac{3}{2}\sqrt[3]{2}$ they have exactly three common points.

b) Each intersection point of the parabola and the hyperbola satisfy the system of equations $y=x^2-a$ and $xy=1$. Hence it satisfy the equations $y^2=x^{2}y-ay$ and $y^2=x-ay$. Subtracting the latter equation from $y=x^2-a$, we obtain the equation $y-y^2=x^2-a-x+ay$, which is equivalent to $(x-\frac{1}{2}{)}^2+(y-\frac{a-1}{2}{)}^2=\frac{(a+1{)}^2+1}{4}$. This is the equation of the circle centered at $(\frac{1}{2};-\frac{a-1}{2})$. Thus, the required locus is the vertical ray $x=\frac{1}{2}$ with the condition $y<-\frac{3}{4}\sqrt[3]{2}+\frac{1}{2}$.