Given curve $y=x^2$ and the another required curve is $y=(x-a)^2$
Intersection point of two curves is
$$
\begin{aligned}
& x^2=(x-a)^2 \\
& x^2=x^2+a^2-2 a x \\
& a(a-2 x)=0 \\
& a=0, x=\frac{a}{2}
\end{aligned}
$$
From $y=x^2$ then,
$$
\mathrm{y}=\frac{\mathrm{a}^2}{4}
$$
Then,intersection point is $\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}^2}{4}\right)$
Take differentiation both sides tothe curves $\mathrm{y}=\mathrm{x}^2$ and $\mathrm{y}$
$$
\begin{aligned}
& =(x-a)^2 \\
& \frac{d x}{d y}=2 x
\end{aligned}
$$
Slope of tangent $m_1=\left.\frac{d y}{d x}\right|_{\left(x=\frac{a}{2}\right)}$
$$
=2 \mathrm{x}=2 \times \frac{4}{2}=\mathrm{a}
$$
Similarly,
Slope of tangent $\mathrm{m}_2=\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(\mathrm{x}=\frac{\mathrm{a}}{2}\right)}$
$$
\begin{aligned}
& \mathrm{m}_2=\left.2(\mathrm{x}-\mathrm{a})\right|_{\left(\mathrm{x}=\frac{\mathrm{a}}{2}\right)} \\
& \mathrm{m}_2=2\left(\frac{\mathrm{a}}{2}-\mathrm{a}\right)=2 \times\left(\frac{-\mathrm{a}}{2}\right)=-\mathrm{a}
\end{aligned}
$$
Now,
$$
\tan \theta=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|=\left|\frac{\mathrm{a}+\mathrm{a}}{1+\mathrm{a} 2}\right|=\frac{2|\mathrm{a}|}{\left|1-\mathrm{a}^2\right|}
$$
so, option (b) iscorrect.