There are two ways. One given slope of tangent and two at a given point on parabola.
One #-# Let the tangent be #y=mx+c#. Normally a line cuts the parabola in two distinct points. To find them let us put this value in the equation of parabola and we get
#mx+c=-1/8x^2+8#
or #x^2+8mx+8c-64=0#
Now for #y=mx+c# to be a tangent, we should have only one root of #x#, which is posiible when discriminant is #0# i.e.
#64m^2-4m(8c-64)=0#
and hence given slope #m# of the tangent, we can find value of #c# to get the equation of tangent #y=mx+c#.
Two #-# Let the tangent to be found on point #(x_0,y_0)#. Note that #y_0=-1/8x_0^2+8#.
As #y=-1/8x^2+8#, slope is given by #(dy)/(dx)=-1/4x# and slope at #x_0# is #-1/4x_0#
and equation of tangent is #y-y_0=-1/4(x-x_0)#
or #y-(-1/8x_0^2+8)=-1/4x_0(x-x_0)#
or #y+1/8x_0^2-8=-1/4x_0x+1/4x_0^2#
or #8y+2x_0x=x_0^2+64#
