<s>[INST] The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through The tangent line to the curve of intersection of the surface $x^2+y^2=z$ and the plane $x+z=3$ at the point $(1,1,2)$ passes through
(A)$(-1,-2,4)$
(B)$(-1,4,4)$
(C)$(3,4,4)$
(D)$(-1,4,0)$
I can find the equation of the tangent line, If I can find the suitable parametrization($\vec{R(t)}$) using the equation $w(\lambda)=\vec{R(t)}|_{(1,1,2)}+ R'(t)|_{(1,1,2)}\lambda$
I made a parametrization by choosing $x=t,z=3-t,y^2=t-t^2.$ But it is not differnetiable at that point. How do I find the suitable parametrization?
 [/INST] Two surfaces, paraboloid $z=x^2+y^2$ and plane $z=3-x$, intersect on ellipse:
$$x^2+y^2=3-x \iff x^2+x+y^2=3 \iff (x+\frac{1}{2})^2-\frac{1}{4}+y^2 = 3 \iff (x+\frac{1}{2})^2+y^2=\frac{13}{4} \\$$
$$\iff \frac{(x+\frac{1}{2})^2}{\frac{13}{4}}+\frac{y^2}{\frac{13}{4}}=1$$
Now the tangent on the intersection curve at the point $P=(1,1,2)$ is easy to find, as well as it is easy to check which one of 4 points lie on this tangent...
Figure.

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