We have, equation of parabola $y=x^{2}$ Let $P\left(\alpha, \alpha^{2}\right)$ is a point on the parabola,
$\therefore$
$y-\alpha^{2}=2 \alpha(x-\alpha)$
$\because$
$\left[\because \frac{d y}{d x}=2 x \Rightarrow \frac{d y}{d x\left(\alpha, a^{2}\right)}=2 \alpha\right]$
$\Rightarrow \quad y=2 \alpha x-\alpha^{2}$
Also, given $y=-x^{2}+4 x-4$
$\therefore-x^{2}+4 x-4=2 a x-a^{2}$
$\Rightarrow \quad x^{2}+2 x(\alpha-2)+\left(4-a^{2}\right)=0$
Discriminant $=0$
$4(\alpha-2)^{2}-4\left(4-a^{2}\right)=0$
$\Rightarrow$
$(\alpha-2)^{2}-\left(4-\alpha^{2}\right)=0$
$\Rightarrow \quad \alpha^{2}-4 \alpha+4-4+\alpha^{2}=0$
$\alpha^{2}-2 \alpha=0$
$\alpha=0, \alpha=2$