$\text{ If one end of a focal chord }\mathrm{AB}\text{ of the parabola }{y}^2=8x\text{ is at }A\left(\frac{1}{2}-2\right),\text{ then the equation of }$

$\text{ the tangent to it at }\mathrm{B}\text{ is : }$

$\begin{array}{l}\left(a{t}_1^2,2a{t}_1\right)=\left(\frac{1}{2},-2\right)\\ 4a=8, a=2, 2a{t}_1=-2, t_1=-\frac{1}{2}, t_1{t}_2=-1, t_2=2\\ \left(a{t}_1^2,2a{t}_2\right)=(8,8), y^2=8x, 2y\frac{dy}{dx}=8, \frac{dy}{dx}=\frac{8}{2y}\\ \frac{dy}{dx}(8,8)=\frac{8}{2\left(8\right)}=\frac{1}{2} ,(y-8)=\frac{1}{2}(x-8), 2y-16=x-8, -8=x-2y, x-2y+8=0\end{array}$