Thread: Points on elliptic paraboloid

1. Points on elliptic paraboloid

Find all points on the elliptic paraboloid $z=x^2+2y^2-1$ at which the normal line to the tangent plane coincides with the line joining the origin to the point (or, equivalently the normal vector to the tangent plane is parallel to the vector from the origin to the point).

I honestly do not where to begin. I am suppose to answer and explain each step.

2. Originally Posted by papabear
Find all points on the elliptic paraboloid $z=x^2+2y^2-1$ at which the normal line to the tangent plane coincides with the line joining the origin to the point (or, equivalently the normal vector to the tangent plane is parallel to the vector from the origin to the point).

I honestly do not where to begin. I am suppose to answer and explain each step.
the normal vector to the tangent plane at (x,y,z) is $\nabla{f}(x,y,z),$ where $f(x,y,z)= x^2+2y^2-z-1.$ thus: $\nabla{f}(x,y,z)=(2x,4y,-1).$ the vector from the origin to the point (x,y,z) is (x,y,z).

so we want to find all points on the surface that satisfy the equation $\nabla{f}(x,y,z)=k(x,y,z),$ for some $k \in \mathbb{R}.$ that gives us: $(2x,4y,-1)=(kx,ky,kz).$ thus we just need to solve the following

system of equations: $2x=kx, \ \ 4y=ky, \ \ -1=kz, \ \ z=x^2+2y^2-1.$ i hope you can take it from here! you should consider the cases $x = 0$ and $x \neq 0$ separately. similarly for $y.$ you will get 5

points that satisfy the above system of equations.

3. How would I solve for k and get all the points??

4. It is a paraboloid. Think of it, can 5 points be at the same length from the origin. Look the attached graphs.

5. Its hard to tell from the graphs... but how would i get the points ? we didnt go over lagrange multipliers... so i cant use that.. how do i solve the system of equations NonCommAlg gives...

6. Originally Posted by papabear
how do i solve the system of equations NonCommAlg gives...
if $x \neq 0,$ the first equation gives you k = 2 and the second and the third give you $y = 0$ and $z = \frac{-1}{2}$ respectively. thus from the fourth equation we have: $x = \pm \frac{\sqrt{2}}{2}.$

if $x =0,$ and $y=0,$ then the fourth equation gives you: $z = -1.$

finally if $x=0$ and $y \neq 0,$ then the second equation gives you k = 4 and the third equation will give us $z = \frac{-1}{4}$ and thus from the fourth equation we get: $y=\pm \frac{\sqrt{6}}{4}.$

so your five points are: $P_1= (\frac{\sqrt{2}}{2},0,\frac{-1}{2}), \ \ P_2=(\frac{-\sqrt{2}}{2},0,\frac{-1}{2}), \ \ P_3=(0,0,-1), \ \ P_4=(0, \frac{\sqrt{6}}{4}, \frac{-1}{4}),$ and $P_5=(0, \frac{-\sqrt{6}}{4}, \frac{-1}{4}).$

7. Maybe these graphs are easier to understand.
But the other guy was right. There are 5 such point sorry. Shortcut are not always shorter.