$\text{ The focal chord of the parabola }(y-2{)}^2=16(x-1)\text{ is a tangent to the circle }$

$x^2+y^2-14x-4y+51=0,\text{ then slope of the focal chord can be }$

$\text{ Given parabola is }(y-2{)}^2=16(x-1)$

$\begin{array}{l}\text{ Here vertex }=(h,k)=(1,2)\\ 4a=16\Rightarrow a=4\\ \text{ Focus }=(h+a,k)=(1+4,2)=(5,2)\\ \text{ The equation of focal chord is }y-2=m\left(x-5\right)\\ \Rightarrow mx-y+2-5m=0\dots \dots \dots ...\left(1\right)\end{array}$

$\text{ Given circle equation is }{x}^2+y^2-14x-4y+51=0\dots \dots \dots ...\left(2\right)$

$\begin{array}{l}\text{ Hence centre }(7,2)\\ \text{ And radius, }r=\sqrt{49+4-51}=\sqrt{2}\end{array}$

Since equation (1) is a tangent of circle equation (2) then

r = d

$\begin{array}{l}\sqrt{2}=\frac{|7m-2+2-5m|}{\sqrt{m^2+1}}\\ (\because d=\text{ perpendiculardistance from centre to equation }(1\left)\right)\end{array}$

$\begin{array}{r}\Rightarrow \sqrt{2}\sqrt{m^2+1}=\left|2m\right|\\ \text{ Squaring on both sides }\end{array}$

$\begin{array}{l}\Rightarrow 2m^2=m^2+1\\ \Rightarrow m=\pm 1\\ \therefore \text{ Slope of focal chord is }1\end{array}$