Write the parabola in standard form

Given parabola: $y^2 = 16x$

Compare with the standard form: $y^2 = 4ax$

So, $4a = 16 \implies a = 4$

Hence the focus is: $(a,0) = (4,0)$

Check that the given point lies on the parabola

Given one end of the focal chord is $(1,4)$.

Substitute into the parabola: $y^2 = 4^2 = 16, \quad 16x = 16(1)=16$

So $(1,4)$ lies on the parabola.

Use the parametric form of the parabola

For $y^2 = 4ax,$ a point on the parabola can be written as: $(at^2, 2at)$

Here $a=4$, so the parametric point is: $(4t^2, 8t)$

Match with $(1,4)$: $8t = 4 \implies t = \frac{1}{2}$

Check x-coordinate: $4t^2 = 4\left(\frac{1}{2}\right)^2 = 1$

Correct.

Use the condition for a focal chord

For the parabola $y^2 = 4ax,$ if one end of a focal chord corresponds to parameter $t$, then the other end corresponds to parameter: $-\frac{1}{t}$

Since $t = \frac{1}{2},$ the other end has parameter: $-\frac{1}{t} = -2$

Find the coordinates of the other end

Using parametric form $(4t^2, 8t)$ with $t=-2$: $x = 4(-2)^2 = 16, \quad y = 8(-2) = -16$

So the other end is: $(16,-16)$

Find the length of the focal chord

Distance between $(1,4)$ and $(16,-16)$ is: $\sqrt{(16-1)^2 + (-16-4)^2}$ $= \sqrt{15^2 + (-20)^2}$ $= \sqrt{225 + 400}$ $= \sqrt{625}$ $= 25$

Compare with options

The length of the focal chord is: $25$

So the correct option is C.