Finding parametric eqution for the tangent line 
Let $C$ be the curve through $P\left(-2,1,5\right)$
  intersecting the surfaces $\Sigma_1:z=x^2+y^2$ and
  $\Sigma_2:z=13-(x^2+4y^2)$. Find the tangent line of $C$ at $P$.

My teacher used the total derivative of each surface to find the orthogonal vector. The technique is simple, I just don't fully understand why she did so. Can anyone please provide a geometric explanation? Thanks!
 A: The total derivative of a function
$f:\mathbb{R}^3\rightarrow\mathbb{R}$
at a point $P=\mathbf{x}_0$ is a linear function
$df:\mathbb{R}^3\rightarrow\mathbb{R}$,
$$
df
=\frac{\partial f}{\partial x}dx
+\frac{\partial f}{\partial y}dy
+\frac{\partial f}{\partial z}dz,
$$
i.e. so that $df(\Delta x,\Delta y,\Delta z)
=\frac{\partial f}{\partial x}\Delta x
+\frac{\partial f}{\partial y}\Delta y
+\frac{\partial f}{\partial z}\Delta z$.
It gives you the linear approximation
$$
f(\mathbf{x}) \approx f(\mathbf{x}_0)
+\frac{\partial f(\mathbf{x}_0)}{\partial x}\Delta x
+\frac{\partial f(\mathbf{x}_0)}{\partial y}\Delta y
+\frac{\partial f(\mathbf{x}_0)}{\partial z}\Delta z
$$
where
$(x,y,z)=\mathbf{x}=\mathbf{x}_0+\Delta\mathbf{x}$
and
$\Delta\mathbf{x}=(x-x_0,y-y_0,z-z_0)$,
and its level sets are planes in $\mathbb{R}^3$
perpendicular to the gradient $\nabla{f}(\mathbf{x}_0)$.
In our case, if we take the cross product of the gradients
at a point of common intersection $\mathbf{x}_0$,
we will also find the tangent to the curve (not line!) $C$,
since this gives us something normal to both gradients and
therefore tangent to $\Sigma_1$ and $\Sigma_2$.
Now $\Sigma_1$ is a paraboloid with global minimum at the origin
and axis of symmetry/revolution along the $z$ axis, and $\Sigma_2$
is similar but inverted, with parabolic vertical cross sections
and elliptic horizontal cross sections and global maximum at $(0,0,13)$. To calculate these gradients, we need to write the equations for these curves as functions:
$$f_1=x^2+ y^2-z\quad\implies\quad\nabla{f_1}=(2x,2y,-1)$$
$$f_2=x^2+4y^2+z\quad\implies\quad\nabla{f_2}=(2x,8y,+1)$$
(the curves will then be the level sets for the values $f_1=0$
and $f_2=13$). This would give us a tangent vector parallel to
$$\nabla{f_1}\times\nabla{f_2}=(10y,-4x,12xy)$$
and hence, at $P(-2,1,5)$, a parametric tangent line
(with parameter $s$ being distance from $P$) of
$$\mathbf{x}(s)=(-2,1,5)\pm\frac{s}{\sqrt{185}}(5,-4,12).$$
Note also that curve $C$ is the locus of points $(x,y,z)$ satisfying
$x^2+y^2=z=13-(x^2+4y^2)$ $\implies$
$2x^2+5y^2=13$ $\implies$
$$
\left(\frac{x}{\sqrt{13/2}}\right)^2+
\left(\frac{y}{\sqrt{13/5}}\right)^2=
13
$$
which could therefore be parametrized by
$$
\mathbf{x}(t)=
\left(
\sqrt{\frac{13}2}\cos{t},
\sqrt{\frac{13}5}\sin{t},
\frac{13}{10}
\left(5\cos^2t+3\sin^2t\right)
\right),
$$
giving yet another way to compute the tangent!
