Answer:
[tex]x=-1-18 t,y=1+19 t,z=2-2 t[/tex]
Step-by-step explanation:
We are given that
Equation of paraboloid
[tex]z=x^2+y^2[/tex]
[tex]z-x^2-y^2=0[/tex]
And equation of the ellipsoid
[tex]3x^2+2y^2+4z^2-21=0[/tex]
We have to find the parametric equation for tangent line to the curve of the intersection of the paraboloid and ellipsoid at point (-1,1,2).
We have to find the normal at point (-1,1,2)
[tex]N_1=-2x\hat{i}-2y\hat{j}+\hat{k}[/tex]
Normal at point (-1,1,2)
[tex]N_1=<2,-2,1>[/tex]
[tex]N_2=6x\hat{i}+4y\hat{j}+8z\hat{k}[/tex]
Normal at point (-1,1,2)
[tex]N_2=<-6,4,16>[/tex]
We rescale and set [tex]N_2=<-3,2,8>[/tex]
The tangent vector to the curve of intersection is given by
[tex]N_1\time N_2[/tex]=[tex]\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-2&1\\-3&2&8\end{vmatrix}[/tex]
[tex]N_1\times N_2=<-18,19,-2>[/tex]
Hence, the tangent line is given by
[tex]x=-1-18 t,y=1+19 t,z=2-2 t[/tex]
