The area of the triangle formed by the tangent and normal drawn to the curve y² = 4x  at (1,2) with Y-axis is (in sq.units)

given curve is $y^2=4x$
Differentiating w.r.t to x 
$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}$ 
Given point P(1,2)
Slope of the tangent at P= $m={\left(\frac{dy}{dx}\right)}_{P(1,2)}=\frac{2}{2}=1$
Area of the required triangle = $\frac{{x_1}^2(1+m^2)}{2\left|m\right|}=\frac{1(1+1)}{2}=\frac{2}{2}=1$