Only $p=2$.

Let $A=(a,a^2),B=(b,b^2)$ $a,b$ are roots of $px^2+qx+r=x^2 \to a+b=-\frac{q}{p-1}, ab=\frac{r}{p-1}$ Tangent in $A$ is $y=2ax-a^2$, tangent in $B$ is $y=2bx-b^2$ so $C=(\frac{a+b}{2}, ab)$ and so $p (\frac{a+b}{2})^2+q\frac{a+b}{2}+r=ab$ or $p(a+b)^2+2q(a+b)+4r=4ab \to p(a+b)^2-2(p-1)(a+b)^2+4(p-1)ab=4ab \to (2-p)( a-b)^2=0$ so $p=2$