I have the surface $z=x^2-y^2$.
I need to find all points on the surface where the normal vector passes through the point $(0,0,2)$.
I started with $\nabla F = \langle -2x,2y,1 \rangle$
Then, since the normal vector needs to pass through $(0,0,2)$ and $(x,y,z)$ then
$(-2x,2y,1) = \lambda(x,y,z-2)$
So I get $\lambda = \pm{1\over 2}$
I end up getting $z={3\over 2}$ and $z={5\over 2}.$
When $z={5\over 2}$, it appears to overshoot the point. It also appears that valid points only occur along the line $y=0$, so is it safe to assume that because this came from the relationship to $y$ it can be thrown out?
Therefore I get $x = \pm \sqrt{\frac {3}{2}}$ and therefore $(\sqrt{\frac {3}{2}},0,{3\over 2})$ and $(-\sqrt{\frac {3}{2}},0,{3\over 2})$.
Looking at the graph I should also get the point where $z=0$ and I'm not sure where that point gets left out and if I'm missing any others.
Is this the correct approach?
Best Answer
At the point $(x,y,z)$ the gradient (the normal vector) is
$ n = (2 x, - 2 y, -1) $
The equation of the normal line is
$P(t) = (x, y, z) + t (2 x, -2 y, -1) $
Setting this equal to $(0,0,2)$ results in
$ (1 + 2 t) x = 0 , (1 - 2 t) y = 0, z = 2+t$, and in addition we have
$ z = x^2 - y^2 $
If $x =0, y=0$ then we have the point $(0,0,0)$.
If $x = 0, y \ne 0$ then $t = \dfrac{1}{2},$ and $z = \dfrac{5}{2} $ , however, $y^2 = x^2 - z $ which has no real solutions.
If $x \ne 0, y = 0 $ then $ t= -\dfrac{1}{2}, z = \dfrac{3}{2} ,x^2 = z + y^2 $ , for which the solutions are $x = \pm \sqrt{\dfrac{3}{2}} $
If $x \ne 0 , y \ne 0$ , then there is no value of $t$ that will satisfy first two equations. Hence no more solutions.
Hence, there are three points $(0,0,0), (\sqrt{\dfrac{3}{2}} , 0, \dfrac{3}{2} ) , (-\sqrt{\dfrac{3}{2}} , 0, \dfrac{3}{2} )$