Prove that mapped tangent plane to the conic section is also a tangent plane to the mapped conic section Let S be a conic section in $P(\mathbb R^3)$. Let $p$ be a projectivity $p: P(\mathbb R^3) \rightarrow P(\mathbb R^3)$, written as $p(x) = Qx$, where $Q$ is orthogonal $3 \times 3$ matrix. We define $L_1 = \{(x,y,z) ~|~ \langle\vec{n} , (x,y,z)\rangle = 0\}$, where $\vec{n}$ is some $3 \times 1$ vector, to be a tangent plane to $S$.
Prove that $L_2 = \{(x, y, z) ~|~ \langle Q\vec{n} , (x,y,z)\rangle = 0\}$ is a tangent plane to the conic section $p(S)$.
Can someone give me a hint on how to solve this problem.
 A: Intuitively, orthogonal transformations preserve lengths and angles, so they also preserve orientations of objects in $3$D space.
More rigorously, let $S$ be defined by the equation $f(x, y, z) = 0$ and let $\vec{x}_0$ be the point where the tangent plane intersected $S$. With the convention that gradients are row vectors, note that $\vec{n} = \nabla f(\vec{x}_0)^T$ and that $p(S)$ is defined by the equation $$g(x, y, z) = f(Q^{-1} (x, y, z)) = f(Q^T (x, y, z)) = 0$$
(Why?) From here, recall that the tangent plane at the point $Q \vec{x}_0$ is given by
$$\langle \nabla g (Q \vec{x}_0)^T, (\vec{x} - Q\vec{x}_0) \rangle = 0$$
But notice that from the chain rule we have $\nabla g (\vec{x}) = \nabla f (Q^T \vec{x}) \cdot Q^T$, and hence the tangent plane at $Q \vec{x}_0$ is given by
\begin{align*}
\langle [\nabla f (Q^T Q \vec{x}_0) \cdot Q^T]^T, (\vec{x} - Q\vec{x}_0) \rangle 
&= \langle Q \nabla f(\vec{x}_0)^T, (\vec{x} - Q\vec{x}_0) \rangle = \langle Q \vec{n}, (\vec{x} - Q\vec{x}_0) \rangle = 0
\end{align*}
But since $\vec{x}_0$ is on $L_1$, namely $\langle \vec{n}, \vec{x}_0 \rangle = 0$, it follows that $\langle Q \vec{n}, Q \vec{x}_0 \rangle = 0$, so the equation above reduces to
$$\langle Q \vec{n}, \vec{x} \rangle = 0$$
which is exactly the equation of $L_2$. Note that this did not depend on the fact that $S$ is a conic section. $\square$
