Showing that the image of a parabola under an invertible linear map is a parabola A parabola $p$ is determined by a line $m$ and a point $P$ not on that line. Let $T$ be an invertible linear map on the real plane. I want to show that $T(p)$, the image of $p$ under this map, is also a parabola. There is a way to do it using equations, but it is very lengthy and I imagine that there are more elegant methods. In particular, I would like to know how to find the line and point that determine the parabola $T(p)$. Are they just the images of $m$ and $P$?
This is for homework so I appreciate that hints would be given and not full solutions.

EDIT.
Let me add an even shorter argument, more suited to those who are not familiar with solid geometry. My original answer at the end.
The equation of any conic section can be written as
$$
\pmatrix{x&y&1}M\pmatrix{x\\y\\1}=0,
$$
where $M$ is a symmetric $3\times3$ matrix. 
This conic section is a parabola if and only if $\det M\ne0$ and $\det M_{33}=0$, where $M_{33}$ is the submatrix formed by the first two rows and columns of $M$. 
Under a nonsingular linear transformation, whose inverse is represented by a $2\times2$ matrix $L$, the equation of the transformed conic section is then:
$$
\pmatrix{x&y&1}\hat{L}^TM\hat{L}\pmatrix{x\\y\\1}
=\pmatrix{x&y&1}M'\pmatrix{x\\y\\1}=0,
\quad\hbox{where}\quad
\hat{L}=\pmatrix{L & 0 \\ 0 & 1}.
$$
But $\det M'=\det(\hat{L}^TM\hat{L})=(\det L)^2\det M\ne0$ and
$\det M'_{33}=\det(L^TM_{33}L)=(\det L)^2\det M_{33}=0$, 
hence the conic section represented by the new equation is still a parabola.
By the way: this argument shows that also an ellipse and a hyperbola are transformed into an ellipse and a hyperbola respectively, under a nonsingular linear mapping.

ORIGINAL ANSWER.
A parabola is always the intersection between a certain circular cone and plane $(x,y)$, that cone having a generatrix parallel to the plane.
You can extend your 2-D linear map to a non-singular 3-D linear map just adding that $z\to z$: under such a map, the cone transforms into another cone. As the mapping is non-singular, you can be certain that there still exists a generatrix of the transformed cone which is parallel to $(x,y)$ plane: it follows that the intersection between the transformed cone and the plane is still a parabola.
Note that the transformed cone will have, in general, an elliptic normal section. But the intersection between cone and plane is in any case a parabola, if a generatrix is parallel to the plane.
