Concept:

The slope of the line tangent to the curve;

\(\rm {y_2-y_1\over x_2-x_1}=\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)_{at(x_1, y_1)}\) = m

Where (x1, y1) are the point on which tangent touches the curve and (x2, y2) is any other point on the tangent

Equation of the line with slope m and an on line points (x1, y1) is:

\(\rm (y-y_1) = m(x-x_1)\)

Distance of a point (p, q) from a line ax + by + c = 0:

D = \(\rm \left|{ap+bq+c\over\sqrt{a^2+b^2}}\right|\)

Calculation:

Given curve y2 = 16x

Differentiating w.r.t x on both sides

⇒ \(\rm \frac{\mathrm{d} }{\mathrm{d} x}y^2=\frac{\mathrm{d} }{\mathrm{d} x}16x\)

⇒ 2y \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}\) = 16

The point on which tangent touches the curve is (1, 4)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = {8\over 4}\) = 2

⇒ Slope of tangent m = 2

Now equation of the tangent: 

\(\rm (y-y_1) = m(x-x_1)\)

⇒ y - 4 = 2(x - 1) 

⇒ y - 2x - 2 = 0​

So, distance of the origin (0, 0) from normal is;

D = \(\rm \left|{ap+bq+c\over\sqrt{a^2+b^2}}\right|\)

⇒ D = \(\rm \left|{1\times0+(-2)\times0+(-2)\over\sqrt{1^2+(-2)^2}}\right|\) 

⇒ D = \(\boldsymbol{\rm \left|{-2\over\sqrt5}\right| = {2\over\sqrt5}}\)