Equation of tangent and normal for translated conics Suppose I have a translated parabola
$(y-k)^2=4a(x-h)$ 
and I need to find the equation of its tangent, can I just replace $x$ and $y$ in 
$y=mx+a/m$ 
by $(x-h)$ and $(y-k)$? 
similarly can I do that with its equation of normal?
And if not, what will be the correct equations of normal and tangent?

P.S. I would appreciate it if you explained it using the theory of conics rather than calculus as I haven’t studied calculus yet.
What basically happens during translation is the shift of the origin.(0,0) shifts to (h,k).Thus, every y coordinate changes to y-k and x to x-h.Think like this-
"You have a circular rubber-band, and on it you have a net.When you 'move' the 'net' keeping the circle's position under it fixed, you actually shift the origin in terms of coordinate geometry.With reference to the previous position of the net, you have shifted it to a new position, by a distance 'h' horizontally and 'k' vertically.Thus the new origin with respect to the old has 'shifted' to (h,k).And so changes the x and y coordinates, which, too, 'shifts' by h and k...
What basically happens during translation is the shift of the origin.(0,0) shifts to (h,k).Thus, every y coordinate changes to y-k and x to x-h.Think like this-
"You have a circular rubber-band, and on it you have a net.When you 'move' the 'net' keeping the circle's position under it fixed, you actually shift the origin in terms of coordinate geometry.With reference to the previous position of the net, you have shifted it to a new position, by a distance 'h' horizontally and 'k' vertically.Thus the new origin with respect to the old has 'shifted' to (h,k).And so changes the x and y coordinates, which, too, 'shifts' by h and k...