We got various problems in this site asking for similar problems btw an ellipsoid and a plane. What if it's btw an ellipsoid and a paraboloid?
I got the equation of both surfaces: $4x^2 + y^2 + z^2 = 9$ and $z = x^2+y^2$ respectively. Now given the tangent line touches their intersection (as an off-origin $xz$-ellipse but this might be useless) at $(-1, 1, 2)$, how can we find the $\underline{v}$ for the parametric equation of the tangent line $\underline{r}(t) = (-1, 1, 2) + t\underline{v}$? Is there any way we need to utilize the directional derivative of either surface? Or we just use $f(x, y, z)$ and $g(x, y, z)$ to describe both surfaces so that $\underline{v} = \nabla f \times \nabla g$ which I got after simplification as $(5, 8, 6)$? Thanks in advance.
Best Answer
You can parametrize the intersection curve of both surfaces and then find the direction vector of the tangent line. But the easier approach here is what you have written towards the end of the question. Let me try and address your question on how we know that tangent planes will intersect in the tangent line. Please note that tangent lines to level curves of both surfaces passing through point $P (-1, 1, 2)$ will lie in the tangent planes to the surfaces at $P$. As the intersection curve is on both surfaces and passed through $P$, its tangent line at $P$ must be on both tangent planes. A line can only lie in two different planes if both planes intersect and it must be the line that is intersection of both planes.
Now coming to the first approach I mentioned. Given surfaces are,
$z = x^2 + y^2$
$4x^2 + y^2 + z^2 = 9$
At intersection, $x = \pm \frac 1 {\sqrt3} \sqrt{9 - z - z^2}, y = \pm \frac 1 {\sqrt3} \sqrt{4z + z^2 - 9}, z = z$
Looking at $P$, we parametrize the intersection curve for $x \leq 0, y \geq 0$.
$ \displaystyle r(t) = \left( - \frac 1 {\sqrt3} \sqrt{9 - t - t^2}, \frac 1 {\sqrt3} \sqrt{4t + t^2 - 9}, t \right)$
Point $P$ is represented by $t = 2$.
$ \displaystyle r'(t) = \left(\frac {2t + 1} {2 \sqrt{27 - 3t - 3t^2}}, \frac {2t + 4} {2 \sqrt{12t + 3t^2 - 27}}, 1 \right)$
$ \displaystyle r'(2) = \left(\frac 56, \frac 86, 1\right)$
So we can take the direction vector of the line as $(5, 8, 6)$ and equation of the tangent line to the intersection curve at $P$ is,
$l: (-1, 1, 2) + s ~(5, 8, 6)$