Question

Consider the ellipsoid 4x^{2}+2y^{2}+z^{2} = 21.

a) The implicit form of the tangent plane to this ellipsoid at (-1,2,3) is

b) The parametric form of the line through this point that is perpendicular to that tangent plane is L(t) =

Answer 1

So I got dz/dx=-8x/2z and dz/dy=-4y/2z but I'm not really sure how to use this information to answer the question

Answer 2

My calc 3 is rusty, so the implicit form for a plane at a point of intersection is in the form a(X-x)+b(Y-y)+c(Z-z)+d=0, if i'm not mistaken we have the required a,b,c and d from the equation of the ellipsoid; furthermore we have the point (x,y,z) given to us so the answer to the first part should be: 4(x--1)^2+2(y-2)^2+1(z-3)^2-21=0.

Answer 3

let see, for a paremetric equation we need a point and a unit vector a point is given (-1,2,3), the perpendicular is a little bit more complex. We know that a plane can be expressed as a point and normal vector, the normal vector (a,b,c) was used in the first portion of the problem and was given to us by the formula for the ellipsoid. That means the normal vector is (4,2,1), using normalization of vectors to make that normal vector into a unit normal vector give (4/4.58, 2/4.58, 1/4.58). we now have all of the information necessary for the paremetric equation l(t)=point +unitnormalvector(t)= (-1,2,3) + <4/4.58, 2/4.58, 1/4.58>t.

Answer 4

Blah blah blah not reading that...