Gradient Vector, Tangent Planes and Normal Lines — Question 4

Question 4 – Solution

Strategy. Parallel tangent planes have parallel normal vectors; impose that condition together with the surface equation.

Step 1: Parallel-gradient condition For $F=x^2+2y^2+z^2$, $$\nabla F=\left\langle 2x,4y,2z\right\rangle.$$ The given plane has normal parallel to $\left\langle 1,-2,1\right\rangle$. Set $$\left\langle 2x,4y,2z\right\rangle=\lambda\left\langle 1,-2,1\right\rangle.$$ Then $x=\lambda/2$, $y=-\lambda/2$, and $z=\lambda/2$. Substitution into the ellipsoid gives $\lambda^2=9$.

Thus the two points are $$\boxed{P_+=\left(\frac 32,-\frac 32,\frac 32\right)},\qquad \boxed{P_-=-P_+}.$$

Step 2: Tangent planes Using the simplified common normal, $$\boxed{x-2y+z=6\quad\text{at }P_+},$$ $$\boxed{x-2y+z=-6\quad\text{at }P_-}.$$

Step 3: Normal lines $$\boxed{\mathbf r=P_++t\left\langle 1,-2,1\right\rangle},\qquad \boxed{\mathbf r=P_-+t\left\langle 1,-2,1\right\rangle}.$$ The equations $\lambda=\pm 3$ exhaust the surface constraint, proving there are no additional points.