We have,
$$
y^2=a x^3+b
$$

Differentiate w.r.t ' $x$ '
$$
\begin{gathered}
2 y \frac{d y}{d x}=3 a x^2 \\
\frac{d y}{d x}=\frac{3 a x^2}{2 y}
\end{gathered}
$$

Slope at $(1,2)$
$$
\begin{aligned}
& \left.\frac{d y}{d x}\right|_{(1,2)}=\frac{3 a(1)^2}{2(2)} \\
& \left.\frac{d y}{d x}\right|_{(1,2)}=\frac{3 a}{4}
\end{aligned}
$$


Given tangent $y=2 x$
Slope of tangent $=2$
From Eqs. (ii) and (iii),
$$
\frac{3 a}{4}=2, a=\frac{8}{3}
$$

Put in Eq. (i),
$$
y^2=\frac{8}{3} x^3+b
$$

Curve passes through $(1,2)$
$$
\begin{aligned}
& & 2^2 & =\frac{8}{3}(1)^3+b \\
\Rightarrow & & b & =4-\frac{8}{3}=\frac{12-8}{3} \\
\Rightarrow & & b & =\frac{4}{3} \\
& \text { So, } & (a, b) & =\left(\frac{8}{3}, \frac{4}{3}\right)
\end{aligned}
$$