A circle of radius 4, drawn on a chord of the parabola ${\mathrm{y}}^2=8\mathrm{x}$ as diameter, touches the axis of the parabola. Then, the slope of the chord is

Given radius of circle =4cm

Given parabola ${\mathrm{y}}^2=8\mathrm{x}$

$$\begin{gathered} \Rightarrow 4\mathrm{a}=8 \\ \Rightarrow \mathrm{a}=2 \\ Let A\left({{\mathrm{at}}_1}^2, 2{\mathrm{at}}_1\right) and B \left({{\mathrm{at}}_2}^2, 2{\mathrm{at}}_2\right) be end points of chord \\ from the given data centre of circle is mid point of AB \\ \Rightarrow (\frac{a(t{{ }_1}^2+{t_2}^2)}{2},\frac{2a(t_1+t_2)}{2}) \\ Since circle touches axis and radius =4 then \\ \frac{2a(t_1+t_2)}{2}=4 \\ \Rightarrow t_1+t_2=2 \\ Now slope of AB\left(m\right)=\frac{2a(t_2-t_1)}{a(t_2-t_1)(t_2+t_1)}=1 \end{gathered}$$