Plotting the diagram of the given data we have,

Now finding the equation tangent at point $\left(1,2\right)$ we get,

$y\times 2=4\left(\frac{x+1}{2}\right)$

$\Rightarrow y=x+1$

Now point on $y$ axis will be point $A\left(0,1\right)$

Now finding the equation of normal we get,

$y-2=-1\left(x-1\right)$

$\Rightarrow y+x=3$

So, point on $y$ axis where normal passes will be $B\left(0,3\right)$

Now area of triangle formed by $\left(1,2\right), \left(0,1\right) \& \left(0,3\right)$ will be,

Area $=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 0& 1& 1\\ 0& 3& 1\end{array}\right|=1$