Thank you for reporting, we will resolve it shortly
Q.
The focal chord to $y^{2}=16 x$ is tangent to $(x-6)^{2}+y^{2}=2$. Then the possible value of the slope of this chord is
Conic Sections
Solution:
For the parabola $y^{2}=16 x$, focus is $(4,0)$. Let $m$ be the slope of focal chord. Then the equation is
$y=m(x-4)$ ... (1)
Given that the above line is a tangent to the circle $(x-6)^{2}$ $+y^{2}=2$ for which the centre is $C(6,0)$ and radius $r$ is $\sqrt{2}$. Therefore, the length of perpendicular from $(6,0)$ to $(1)$ is $r$.
Therefore,
$\frac{|6 m-4 m|}{\sqrt{m^{2}+1}}=\sqrt{2}$
or $ 2 m^{2}=m^{2}+1 $
or $ m^{2}=1 $
or $ m=\pm 1$