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

Question

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?

Answer 1

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.

Answer 2

Let $f(x,y,z) = x^2+y^2-z$ and $g(x,y,z) = x+z-3$. Then $P=(1,1,2)$ lies on the intersection of the surfaces $S_1 = \{f=0\}$ and $S_2 = \{g=0\}$.

The tangent plane to $S_1$ at $P$ is normal to $\mathbf{n}_1 = \nabla f(P) = (2,2,-1)$. The tangent plane to $S_2$ at $P$ is normal to $\mathbf{n}_2 = \nabla g(P) = (1,0,-1)$. The line $L$ through $P$ tangent to $S_1 \cap S_2$ is the intersection of the two tangent planes. That is, $L$ passes through $P$ in the direction $\mathbf{n}_1 \times \mathbf{n}_2$.

This is enough information to parametrize $L$ and decide which of the four points $A$, $B$, $C$, and $D$ lie on it. A quicker strategy may be to find the four vectors $\overrightarrow{PA}$, $\overrightarrow{PB}$, $\overrightarrow{PC}$, and $\overrightarrow{PD}$, and check with the dot product which are perpendicular to both $\mathbf{n}_1$ and $\mathbf{n}_2$. \begin{align*} \overrightarrow{PA} &= (-2,-3,-6) \\ \overrightarrow{PB} &= (-2,3,2) \\ \overrightarrow{PC} &= (2,3,2) \\ \overrightarrow{PD} &= (-2,-3,-2) \\ \end{align*} We see that only $\overrightarrow{PB}$ is perpendicular to $\mathbf{n}_2$. So assuming the correct answer is listed, it would have to be B.

As a further check, we see that $\overrightarrow{PB}$ is perpendicular to $\mathbf{n}_1$. In fact, $\mathbf{n}_1 \times \mathbf{n}_2 = (2,-3,-2) = - \overrightarrow{PB}$.