Identify the parabola and its focus

Given parabola: $y^2=16x$ Comparing with the standard form $y^2=4ax$, we get $4a=16 \implies a=4$ So the focus is $S=(4,0)$

Verify that point $P$ lies on the parabola

Given $P=(1,-4)$ Check: $(-4)^2=16,\, 16(1)=16$ Hence $P$ lies on the parabola.

Find the equation of the focal chord through $P$

Since $PQ$ is a focal chord, it passes through the focus $S=(4,0)$ and point $P=(1,-4)$.

Slope of line $SP$: $m_1=\frac{0-(-4)}{4-1}=\frac{4}{3}$

Equation of the line through $P(1,-4)$: $y+4=\frac{4}{3}(x-1)$ $3y+12=4x-4$ $4x-3y-16=0$

Find the second point of intersection $Q$ with the parabola

From the line, $x=\frac{3y+16}{4}$

Substitute into $y^2=16x$: $y^2=16\cdot \frac{3y+16}{4}$ $y^2=4(3y+16)$ $y^2-12y-64=0$

Solve: $y=\frac{12\pm\sqrt{144+256}}{2}=\frac{12\pm 20}{2}$ So, $y=16 \quad \text{or} \quad y=-4$

Since $y=-4$ corresponds to $P$, for $Q$ we take $y=16$ Then $x=\frac{3(16)+16}{4}=\frac{64}{4}=16$ Hence $Q=(16,16)$

Find the ratio in which the focus divides $PQ$

Let the focus $S=(4,0)$ divide $P(1,-4)$ and $Q(16,16)$ in the ratio $m:n$, i.e. $PS:SQ=m:n$

Using section formula, $\left(\frac{m x_2+n x_1}{m+n},\frac{m y_2+n y_1}{m+n}\right)=(4,0)$ where $(x_1,y_1)=(1,-4)$ and $(x_2,y_2)=(16,16)$.

From the $y$-coordinate: $\frac{16m-4n}{m+n}=0$ $16m-4n=0$ $4m=n$

Therefore, $m:n=1:4$

Since $\gcd(m,n)=1$, we take $m=1,\; n=4$

Compute $m^2+n^2$

$m^2+n^2=1^2+4^2=1+16=17$

Compare with stored answer

Derived answer: $17$

Stored correct answer: A = 17

They match.