Find the equation of the tangent plane and symmetric equation for the normal line to the given surface at P:

Question

xy^2 + zy^2 + 4y -xz^2 = 18
P(-2,0,3)

Answer 1

Define the surface as:
f(x,y,z)=xy^2+zy^2+4y-xz^2-18=0

Then check that P0(-2,0,3) is on the surface by evaluating f(-2,0,3) and confirm that it is zero.

Let fx,fy,fz be the partial derivatives of f(x,y,z) with respect to x,y and z respectively, and evaluated at P0(-2,0,3).

The tangent plane at P is then given by:
fx(x-x0)+fy(y-y0)+fz(z-z0)=0
where (x0,y0,z0) is (-2,0,3).