If one end of a focal chord of the parabola, y ^ 2 = 16x is at (1,...

The given parabola ${y}^{2} = 16x$ is of the form ${y}^{2} = 4ax,$ hence $a = 4 and the focus of the parabola is \left(a, 0\right) = \left(4, 0\right) .$

The equation of a line joining the points $\left({x}_{1}, {y}_{1}\right) and \left({x}_{2}, {y}_{2}\right)$ is $y - {y}_{1} = \left(\dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}\right)\left(x - {x}_{1}\right) .$

Therefore, equation of chord joining $P\left(1, 4\right)$ to focus $S\left(4, 0\right)$ is

$y - 0 = - \dfrac{4}{3}\left(x - 4\right)$

$⇒ 3y = - 4x + 16$

$⇒ 4x + 3y - 16 = 0$

To find the points where this line will cut the parabola put $x = \dfrac{{y}^{2}}{16}$ from the parabola into the line, to get

$4\left(\dfrac{{y}^{2}}{16}\right) + 3y - 16 = 0$

$⇒ {y}^{2} + 12y - 64 = 0$

$⇒ \left(y + 16\right)\left(y - 4\right) = 0$

$⇒ y = - 16$ or $y = 4$

And $x = \dfrac{{y}^{2}}{16}$

$⇒ x = 16$ or $x = 1$

Thus, the points are \left(1, 4\right)$and \left(16, - 16\right),$ but \left(1, 4\right)$is the given point$P .$</p><p>$⇒ Q = \left(16, - 16\right)$</p><p>The distance between the points \left({x}_{1}, {y}_{1}\right) \text{ and } \left({x}_{2}, {y}_{2}\right)$ is $\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

So, the length of focal chord $PQ$ is

$PQ = \sqrt{{\left(16 - 1\right)}^{2} + {\left(- 16 - 4\right)}^{2}}$

$= \sqrt{225 + 400} = \sqrt{625} = 25$ units.