$\text{ Let }\mathrm{P}\text{ be the point on the parabola }{y}^2=4x\text{ which is at the shortest distance from the }$$\text{ centre }\mathrm{S}\text{ of the circle }{x}^2+y^2-4x-16y+64=0\text{ . Let }\mathrm{Q}\text{ be the point on the circle dividing }$$\text{ the line segment SP internally. Then }$

$(x-2{)}^2+(y-8{)}^2=4$

$\text{ Normal to }{y}^2=4x$

$y+xt=2t+t^3$

$\text{ If it passes through }(2,8)\text{ then }8+2\mathrm{t}=2\mathrm{t}+{\mathit{t}}^3$

$\Rightarrow t=2$

$\therefore P(4,4)\text{ eq to }\mathrm{SP}\text{ is }2x+y=12$

$\underset{¯}{\underset{¯}{S}(2,8)}SP=\sqrt{4+16}=2\sqrt{5}$

$\text{ Then }Q\left(\frac{2+4\lambda }{1+\lambda },\frac{8+4\lambda }{1+\lambda }\right)=\left(4-\frac{2}{\lambda +1},4+\frac{4}{\lambda +1}\right)$

Q lie, on the circle

${\left(2-\frac{2}{\lambda +1}\right)}^2+{\left(\frac{4}{\lambda +1}-4\right)}^2=4$

$\Rightarrow \left(4+16\right){\left(1-\frac{1}{\lambda +1}\right)}^2=4$

${\left(1-\frac{1}{\lambda +1}\right)}^2=\frac{1}{5}\Rightarrow \lambda =\frac{\sqrt{5}+1}{4}$

Let Q is (2-t,8+2t)

$\therefore t^2+4t^2=4\Rightarrow t=\pm \frac{2}{\sqrt{5}}$

$\text{ Ratio }\frac{8-8-2t}{8+2t-4}=\frac{-t}{t+2}=\frac{\sqrt{5}+1}{4}$

X-intercept of normal=6

$\text{ Slope of tangent to circle at }\mathrm{Q}=+\frac{1}{2}$