While the slope of tangent at #(x_1,f(x_1))# for #f(x)# is given by value of first derivative at that point i.e. #f((x_1)#,

as normal and tangent are perpendicular to each other (i.e. their product is #-1#), slope of normal is #1/(f'(x_1))#.

Now we have #f(x)=x-sqrt(x^2+4)#, and we are seeking tangent at #x=2# i.e. at #(2,2-sqrt(2^2+4))# i.e. at #(2,2-2sqrt2)#.

Now #f'(x)=1-(2x)/(2sqrt(x^2+4))=1-x/sqrt(x^2+4)#

and slope of tangent is #f'(2)=1-2/(2sqrt2)=1-1/sqrt2=(sqrt2-1)/sqrt2#

and equation of tangent is #y-2+2sqrt2=(sqrt2-1)/sqrt2(x-2)#

i.e. #sqrt2y-2sqrt2+4=(sqrt2-1)x-2sqrt2+2#

or #(sqrt2-1)x-sqrt2y-2=0#

and hence slope of normal is #sqrt2/(1-sqrt2)# and equation of normal is

#y-2+2sqrt2=sqrt2/(1-sqrt2)(x-2)#

or #y(1-sqrt2)-2(1-sqrt2)^2=sqrt2x-2sqrt2#

or #sqrt2x-y(1-sqrt2)-2sqrt2+2(3-2sqrt2)=0#

or #sqrt2x-y(1-sqrt2)-6sqrt2+6=0#

graph{((sqrt2-1)x-sqrt2y-2)(sqrt2x-y(1-sqrt2)-6sqrt2+6)(x-sqrt(x^2+4)-y)=0 [-5, 5, -2.5, 2.5]}