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Let us find a point $(x_0,y_0)$ that lies on all parabolae $P_a$ for any real $a$.

The equation of $P_a$ is $y=x^2+ax+(2014-a)$.

Let us try to find $x_0$ and $y_0$ such that for all $a$, $y_0=x_0^2+a{x}_0+(2014-a)$.

Rewriting: $$y_0=x_0^2+a{x}_0+2014-a$$ $$y_0=x_0^2+2014+a(x_0-1)$$ For this to be independent of $a$, the coefficient of $a$ must be zero: $$x_0-1=0⟹x_0=1$$ So, for $x_0=1$: $$y_0=(1{)}^2+a(1)+2014-a=1+a+2014-a=1+2014=2015$$ Therefore, the point $(1,2015)$ lies on all parabolae $P_a$.

Thus, all parabolae $P_a$ pass through the same point $(1,2015)$.