Find parametric equations for the tangent line to the curve of intersection of the paraboloid

Question

z = x2 + y2
and the ellipsoid
6x2 + 5y2 + 7z2 = 39
at the point
(−1, 1, 2)

Answer 1

if u and v are functions of x,y,z, then the normals are

Nu = ∇u
Nv = ∇v

The tangent line of the intersection is perpendicular to both normals

T(x,y,z) = ∇u × ∇v

∇u = 2xi + 2yj - k
∇v = 12xi + 10yj + 14zk

T(-1,1,2) = 33i + 34j + 2k
so that would be

x = -1 + 33t
y = 1 + 34t
z = 2 + 2t

Answer 2

can you please explain to me where you got the numbers for T(-1,1,2) I know you did the cross product but I can not seem to get the same numbers as you. Please explain

Answer 3

he crossed the partial derivatives

Answer 4

hey nim again,
Find parametric equations for the tangent line to the curve of intersection of the paraboloid z =
x
2 + y
2 and the elliposoid x
2 + 4y
2 + z
2 = 9 at the point P(1, −1, 2).
Comments: A tangent vector to the curve of intersection is given by N1 × N2 where N1 is normal
to the graph of z = x
2 + y
2 at the point P and N2 is normal to the level surface F(x, y, z) =
x
2 + 4y
2 + z
2 = 9 at the point P.
Now N1 = h−2x, −2y, 1i. At the point P, N1 = h−2, 2, 1i. The vector N2 = ∇F = h2x, 8y, 2zi.
At P, N2 = h2, −8, 4i. We rescale and set N2 = h1, −4, 2i.
We compute N1 × N2 = h8, 5, 6i which is tangent to the curve of intersection at the point P.
Hence the tangent line is given by x = 1 + 8t, y = −1 + 5t, z = 2 + 6t.

found this at math(dot)drexel(dot)edu/~rboyer/courses/math200_sum_2012/inclass_week7_comments.pdf
and it helped, hope it helps