Hi all,
the problem is:
Given the paraboloid P= 3x^2 + 2y^2 -2z = 1 and a point A(1,1,2) on it,
find the line perpendicular to P at point A.
I would appreciate any help
I don't have much time now, so I can't walk through the problem with you. here is the methodology though. hopefully it helps.
Step 1:
Find the tangent plane to the surface of the paraboloid at the point (1,1,2). This will be of the form ax + by + cz = d
Step 2:
Use the normal vector of the plane from step 1, this will be the vector <a,b,c>, as the direction vector for the line.
Step 3:
Write your solution. The line has the form:
(1,1,2) + t<a,b,c>
where t is a parameter. Write the line in the required form (the vector form, the symmetric form, or the parametric form)