Gradient Vector, Tangent Planes and Normal Lines — Question 10

Question 10 – Solution

Strategy. If the origin lies on the normal line at $P$, then the vector from $P$ to the origin must be parallel to $\nabla F(P)$.

Step 1: Parallelism condition Write $P=(p,q,r)$. Since $$\nabla F(P)=\left\langle 2p,4q,6r\right\rangle,$$ the origin lies on the normal line exactly when some $t$ satisfies $$\left\langle-p,-q,-r\right\rangle=t\left\langle 2p,4q,6r\right\rangle.$$ If $p\ne 0$, then $t=-1/2$; if $q\ne 0$, then $t=-1/4$; if $r\ne 0$, then $t=-1/6$. One value of $t$ cannot meet two of these conditions, so at most one coordinate is nonzero.

Step 2: Points and planes Using the ellipsoid equation on each coordinate axis gives exactly six points: $$(\pm\sqrt 6,0,0),\qquad(0,\pm\sqrt 3,0),\qquad(0,0,\pm\sqrt 2).$$ Their tangent planes are, respectively, $$\boxed{x=\pm\sqrt 6},\qquad \boxed{y=\pm\sqrt 3},\qquad \boxed{z=\pm\sqrt 2}.$$

Step 3: Normal lines and verification The corresponding normal lines are $$\boxed{\mathbf r=(\pm\sqrt 6,0,0)+t\left\langle 1,0,0\right\rangle},$$ $$\boxed{\mathbf r=(0,\pm\sqrt 3,0)+t\left\langle 0,1,0\right\rangle}, \qquad \boxed{\mathbf r=(0,0,\pm\sqrt 2)+t\left\langle 0,0,1\right\rangle}.$$ These are the coordinate axes written through their stated points. Every one passes through the origin and is perpendicular to its listed coordinate plane. The parallelism argument rules out mixed-coordinate points, while the surface equation rules out the origin itself, so the list is complete.