Suppose that the quadric surface S is given by $z = x^2 + x + 2y^2 + 3y$ and the plane is given by $x + y + z = k$, where k is a constant.
Find the vector equation for the tangent line to the curve of intersection between S and plane when $k=3$ at the point (1, 0, 2).
Find the value of k for which the surface S is tangent to plane. For this plane with the value of k found, find the coordinates of the point of tangency.
For the first part of the question, I managed to derive this:
Equating S and the plane together gives - $x^2+2x+2y^2+4y=3$. Letting $x=t$, I am able to find y in terms of t, $y=1+\sqrt{\frac{5}{2}-\frac{t^2}{2}-t}$ and z in terms of t, $z=2-t-\sqrt{\frac{5}{2}-\frac{t^2}{2}-t}$.
After finding the first derivative $f_x$,$f_y$ and $f_z$, I am able to substitute $x=t=1$ into $f_x$,$f_y$ and $f_z$, which gives me (1, -1, 0). Hence the tangent line is : $r(t) = (1,0,2)+t(1,-1,0)$.
However, for the second part of the question, I am unsure of how to solve using the same method as $k$ is now an unknown and I am only able to derive that the normal to the plane is (1, 1, 1). May I get some help with the second part?