If one end of a focal chord of the parabola, y^2=16x is at (1,4), then the length of this focal chord is

The given parabola ${y}^{2}=16x$y2=16x is of the form ${y}^{2}=4ax,$y2=4ax, hence $a=4$a=4 and the focus of the parabola is $(a,0)=(4,0).$(a,0)=(4,0).

The equation of a line joining the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $y-{y}_{1}=(\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}})(x-{x}_{1}).$y−y1​=(x2​−x1​y2​−y1​​)(x−x1​).

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

$$y-0=-\frac{4}{3}(x-4)$$y−0=−34​(x−4)

$$\Rightarrow 3y=-4x+16$$⇒3y=−4x+16

$$\Rightarrow 4x +3y-16=0$$⇒4x+3y−16=0

To find the points where this line will cut the parabola put $x=\frac{{y}^{2}}{16}$x=16y2​ from the parabola into the line, to get

$$4(\frac{{y}^{2}}{16})+ 3y-16=0$$4(16y2​)+3y−16=0

$$\Rightarrow {y}^{2}+ 12y-64=0$$⇒y2+12y−64=0

$$\Rightarrow (y+16)(y-4)=0$$⇒(y+16)(y−4)=0

$\Rightarrow y=-16$⇒y=−16 or $y=4$y=4

And $x=\frac{{y}^{2}}{16}$x=16y2​

$\Rightarrow x=16$⇒x=16 or $x=1$x=1

Thus, the points are $(1,4)$(1,4) and $(16,-16),$(16,−16), but $(1,4)$(1,4) is the given point $P.$P.

$$\Rightarrow Q=(16,-16)$$⇒Q=(16,−16)

The distance between the points ({x}_{1},{y}_{1})&({x}_{2},{y}_{2}) is $\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}}$(x1​−x2​)2+(y1​−y2​)2​

So, the length of focal chord $PQ$PQ is

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

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