Let the point P of the focal chord PQ of the parabola $$y^2 = 16x$$ be $$(1, -4)$$. If the focus of the parabola divides the chord PQ in the ratio $$m : n$$, $$\gcd(m, n) = 1$$, then $$m^2 + n^2$$ is equal to :

The given parabola is $$y^{2}=16x$$.
Write it in the standard form $$y^{2}=4ax$$ to identify $$a$$.

Comparing, $$4a = 16 \implies a = 4$$.
Hence the focus is $$S(4,0)$$.

For a parabola $$y^{2}=4ax$$, any point can be written parametrically as $$\bigl(at^{2},\,2at\bigr)$$.
With $$a = 4$$, a point corresponding to parameter $$t$$ is $$P(t)\; \equiv\; \bigl(4t^{2},\,8t\bigr)$$.

The point $$P(1,-4)$$ lies on the parabola, so find its parameter:

Set $$4t^{2}=1 \implies t^{2}=\tfrac14 \implies t=\pm\tfrac12$$.
Set $$8t=-4 \implies t=-\tfrac12$$.
Therefore the parameter for point $$P$$ is $$t_{1} = -\tfrac12$$.

Property of a focal chord: if the end-points have parameters $$t_{1}$$ and $$t_{2}$$, then $$t_{1}t_{2} = -1$$.

Hence $$t_{2} = -\dfrac{1}{t_{1}} = -\dfrac{1}{-\tfrac12} = 2$$.

Coordinates of the other end $$Q$$ of the focal chord (using $$t_{2}=2$$):

$$Q \; \equiv\; \bigl(4(2)^{2},\,8(2)\bigr) \;=\; (16,\,16).$$

Let the focus $$S(4,0)$$ divide the chord $$PQ$$ internally in the ratio $$m:n$$ (with $$\gcd(m,n)=1$$).
For internal division of $$P(x_{1},y_{1})$$ and $$Q(x_{2},y_{2})$$ in the ratio $$m:n$$, the coordinates are

$$\biggl(\dfrac{mx_{2}+n x_{1}}{m+n},\; \dfrac{my_{2}+n y_{1}}{m+n}\biggr).$$

Substitute $$P(1,-4)$$, $$Q(16,16)$$ and $$S(4,0)$$:

For the $$x$$-coordinate:

$$4 = \dfrac{m\cdot16 + n\cdot1}{m+n}.$$

Multiply: $$4(m+n) = 16m + n \implies 4m + 4n = 16m + n \implies 12m = 3n \implies n = 4m.$$

For the $$y$$-coordinate (check consistency):

$$0 = \dfrac{m\cdot16 + n(-4)}{m+n} \implies 16m = 4n \implies n = 4m,$$
which matches the previous result.

Thus $$m:n = m:4m = 1:4$$ after cancelling the common factor.
So $$m = 1, \; n = 4$$.

Required value:

$$m^{2}+n^{2} = 1^{2} + 4^{2} = 1 + 16 = 17.$$

Therefore, $$m^{2}+n^{2}=17,$$ corresponding to Option A.