Answer:
Option C
Explanation:
Since, x-y=-4k or y=x+4k is a tangent to the parabola $y^{2}=8x$, therefore
$4k=\frac{2}{1}$ $[\because 4a=8 \Rightarrow a=2]$
$\Rightarrow$ $k=\frac{1}{2}$
Also points of concept p is $\left(\frac{a}{m^{2}},\frac{2a}{m}\right)=(2,4)$
Now, equation of normal to the $y^{2}=8x$ at (2,4) is
$(y-4)=\frac{-4}{2(2)}(x-2)$
[$\because$ Equation of normal to the parabola , $y^{2}=4ax$ at
$(x_{1},y_{1})is y-y_{1}=\frac{-b}{2a}(x-x_{1})$
$\Rightarrow$ $y-4=-1(x-2)$
$\Rightarrow$ $y-4=-x+2$
$\Rightarrow$ $x+y=6$
$\Rightarrow$ $x+y-6=0$
The perpendicular distance of normal from (k,2k) ie $(\frac{1}{2},1)$
$=\frac{|\frac{1}{2}+1-6|}{\sqrt{1^{2}+1^{2}}}=\frac{|\frac{3}{2}-6|}{\sqrt{2}}=\frac{9}{2\sqrt{2}}$
