I've got a problem trying to figure out what I'm doing wrong with these question regarding finding the equation of the tangent plane to a parametrised surface. A surface is parametrised by
$$x = u^2-v^2, \ \ \ \ y = u + v, \ \ \ \ z = u^2 + 4v$$
I know this is the parametrisation $$\Phi(u,v) = (u^2-v^2, \ u+v,\ u^2+4v)$$ Also, the surface is not smooth at $(u,v) = (2,-2)$
My problem is:
What is the equation of the tangent plane to the surface at $\Big (\frac{-1}{4},\frac{1}{2},2 \Big) ?$
So far, I've done the following:
I set $$(u^2-v^2, \ u+v,\ u^2+4v) = \Big (\frac{-1}{4},\frac{1}{2},2 \Big)$$
So the system of equations becomes: $$ (1)\ \ \ u^2 - v^2 = \frac{-1}{4}$$ $$ (2)\ \ \ u + v = \frac{1}{2}$$ $$ (3)\ \ \ u^2 + 4v = 2$$
$(3) - 4 (2)$ $$\implies u^2 - 4u = 2-4(\frac{1}{2})$$ $$u^2-4u = 0$$ $$u = 0$$ or $$u = 4$$
Putting $u=0$ into (2), we get $v=\frac{1}{2}$, which is the correct and apparently only solution set $(u,v) = (0,\frac{1}{2})$.
But we also have $u=4$ which, when put into (2), gives another (incorrect) solution set of $(u,v) = (4,\frac{-7}{2})$
My question is:
What exactly have I done wrong?
Cheers guys :)
Giving
$$ S(u,v) = (u^2-v^2,u+v,u^2+4v)\\ p = (x,y,z)\\ p_0 = \left(-\frac 14,\frac 12,2\right) $$
Solving
$$ S(u,v) = p_0 \Rightarrow \left(u = 0, v = \frac 12\right) $$
now calling
$$ \vec t_u = S_u = (2u,1,2u)\\ \vec t_v = S_v = (-2v,1,4)\\ \vec n = \vec t_v\times\vec t_u = 2(u-2,4u+2u v, -u-v) $$
at $p = p_0$ we have $\vec n = \vec n_0 = (-4,0,-1)$
and the tangent plane is
$$ (p-p_0)\cdot \vec n_0 = 0\Rightarrow z-4x+1=0 $$
Attached a plot showing the tangency and $\vec n_0$ in red.