Find Find parametric equations for the tangent line to the curve of intersection of $z=x^2+y^2$ and $6x^2+5y^2+3z^2 =23$ at $(−1, 1, 2).$
I tried plugging in $z=x^2+y^2$ in the second equation to obtain $6x^2+5y^2+3(x^2+y^2)^2=23$, and then converting it to polar to obtain $r^2\left(6\cos^2\theta+5\sin^2\theta\right)+3r^4=23$. I'm not sure to go from here. Can anyone show me how to find the intersection between the surfaces?
Hint: the line tangent to the intersection curve is tangent to both surfaces, hence orthogonal to the gradients of the defining functions. The line is:
$$\mathbf{X}(t) = (-1,1,2) + t (\nabla f(-1,1,2) \times \nabla g(-1,1,2)), \quad t \in \Bbb R,$$ where $f(x,y,z) = x^2+y^2 - z$ and $g(x,y,z) = 6x^2+5y^2+3z^2$.