Showing that the image of a parabola under an invertible linear map is a parabola

Question

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.

Answer 1

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.

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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.

Answer 2

This is a (maybe unexpected) explanation using the so-called quadratic Bezier representation of parabolas.

(classical) result: Any parabola can be given the following parametric representation:

$$\binom{x}{y}=(1-t)^2\binom{x_A}{y_A}+2t(1-t)\binom{x_B}{y_B}+t^2\binom{x_C}{y_C}.$$

(Convert segment of parabola to quadratic bezier curve) (https://en.wikipedia.org/wiki/B%C3%A9zier_curve)

Symbolically:

$$\tag{1} M=(1-t)^2 A+2t(1-t)B+t^2 C.$$

where $A$ and $C$ are any distinct points of the parabola, and intermediate point $B$ is any point not aligned with $A$ and $B$ (in fact, it is the point of intersection of tangents in $A$ and $C$). Let us denote this parametric curve by ${\frak B}(A,B,C)$ : it is the Bezier curve defined by points $A,B,C$.

Now, applying any (invertible) linear transformation $L$ to (1) gives:

$$\tag{2} L(M)=(1-t)^2 L(A)+2t(1-t)L(B)+t^2 L(C),$$

(2) means that the image of Bezier curve ${\frak B}(A,B,C)$ by linear transformation $L$ is Bezier curve ${\frak B}(L(A),L(B),L(C))$, thus a parabola.

Remarks:

1) I leave as an exercise the fact that $L(A),L(B),L(C)$ aren't aligned (the fact that $L$ is invertible play here a role).

2) The sum of the weights in relationship (1) is always 1:

$$\forall t \in \mathbb{R} \ \ \ \ \ (1-t)^2 +2t(1-t)+t^2=1 $$

3) More generally, a linear transform preserves weighted sums (barycenters): the image of a weighted sum of vectors is the weighted sum of their images (with the same coefficients).

Answer 3

You mean the focus $P$ and the directrix $m$? They determine parabola $p$ using locus of equidistant points. Unfortunately, in general, length ratios are not preserved by linear maps. However, in the Wikipedia article Parabola is mention of dual parabola which may help in your proof. There are other possible properties that you may want to use in the article as well.

Answer 4

Hint: You can assume without loss generality that the parabola $P$ is $\stackrel{\sim}{P}\,:y=x^2$, since any parabola can be taken to $\stackrel{\sim}{P}$ via a combination of rotation, translation, and scaling.

Hence, it suffices to consider the image of $(x,x^2)$ under invertible $T$.

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It is hard to give a hint that does not solve the problem, but I encourage you to reflect on the 'without loss of generality' consideration above and think about how one can express $T$ in terms of simpler transformations for which you know the desired property holds, individually.