Ans: (B)
Hint : \(y=x^{2}\)
\(P\left(\alpha, \alpha^{2}\right)\) is a point on this parabola.
$$
\begin{aligned}
&\therefore \mathrm{y}-\alpha^{2}=2 \alpha(\mathrm{x}-\alpha) \\
&\mathrm{y}=2 \alpha \mathrm{x}-\alpha^{2} \ldots(1) \text { is a tangent } \\
&\therefore 2 \alpha \mathrm{x}-\alpha^{2}=-\mathrm{x}^{2}+4 \mathrm{x}-4 \\
&\mathrm{x}^{2}+2 \mathrm{x}(\alpha-2)+\left(4-\alpha^{2}\right)=0
\end{aligned}
$$
$$
\begin{aligned}
&\text { Discriminant }=0 \\
&4(\alpha-2)^{2}-4\left(4-\alpha^{2}\right)=0 \\
&\alpha^{2}-4 \alpha+A-A+\alpha^{2}=0 \\
&\alpha^{2}-2 \alpha=0 \\
&\alpha=0, \alpha=2
\end{aligned}
$$
