# How to find the symmetric equations of normal line at point (1, -1, 4) in the plane if z=f(x,y)=(4x^3)(y^4)??

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You need to find the equation of tangent plane to the surface `z=f(x,y)=(4x^3)(y^4)` at the point (1,-1,4), hence you need to determine the partial derivatives such that:

`f_x = (del f)/(del x) = 12x^2*y^4` (f is differentiated with respect to x only)

`f_y = (del f)/(del y) = 16y^3*x^3` (f is differentiated with respect to y only).

It follows that `f_x (1,-1) = 12(1)^2*(-1)^4 = 12` and`f_y (1,-1) =16(-1)^3*1^3 = -16`

Notice that the tangent plane to the surface at the point (1,-1,4) is:

`z - 4 = f_x (1,-1)*(x-1) + f_y (1,-1)*(y+1)`

`z - 4 = 12(x-1) - 16(y+1)`

Opening the brackets yields:

`z - 4 - 12x + 12 + 16y + 16 = 0`

`-12x + 16y + z + 24 = 0 =gt 12x - 16y - z - 24 = 0`

Hence, evaluating the symmetic equation yields:

`(x-1)/(-12)=(y+1)/16=(z-4)/1`

`` **Hence, evaluating the symmetric equation of normal line to surface z, at the point (1,-1,4), yields `(x-1)/(-12)=(y+1)/16=(z-4)/1.` **