Question

Find parametric equations for the tangent line to the curve of intersection of the paraboloid z=x^2 + y^2 and the ellipsoid 4x^2 + y^2 + z^2 = 9 at the point (-1,1,2).

Answer 1

What I have so far:
The curves will intersect when:\[4x^2 +y^2 +(x^2+y^2)^2=9\] This gives:\[4x^2+y^2+x^4+2x^2y^2+y^4=9\]If x=f(t), y=g(t)then we have:\[8x(dx/dt)+2y(dy/dt)+4x^3(dx/dt)+4xy^2(dx/dt)
\]\[+4x^2y(dy/dt)+4y^3(dy/dt)=0\]Inputting our point x=-1, y=1 we get a relation between (dx/dt) and (dy/dt):\[-8(dx/dt)-4(dx/dt)-4(dx/dt)=-2(dy/dt)-4(dy/dt)-4(dy/dt)\]and,\[16(dx/dt)=10(dydt)\]

Answer 2

The gradient of the curve of intersection is given by:\[<8x+4x^3+4xy^2,2y+4x^2y+4y^3>\]The gradient vector at the given point is:\[<-8-4-4,2+4+4>=<-16,10>\]Also, the dot product between the gradient vector and any tangent vector is zero so:\[<-16,10>*<x+1,y-1>=0\]This gives:\[-16x-16+10y-10=0\]Solving we get:\[10y=16x+26\]and\[y=(16/10)x+(26/10)\]So our parametric equation for y is:\[y=y _{0}+(dy/dt)t\]The above gives:\[(16/10)(x)+(26/10)=1+(16/10)(dx/dt)t\]and,\[16x+26=10+16(dx/dt)t\]Solving for x gives us a parametric equation of a line in x:\[16x=-16+16(dx/dt)t\]Thus,\[x=-1+(dx/dt)t\]And i've crashed into a wall. The book gives, for the parametric equation of x:\[x=-1-10t\]But I cannot get this, try as I might!