How can we be sure that a dilation of a hyperbola is still a hyperbola?

Question

Paul lockhart in his book "Measurement" shows that every hyperbola is a dilated version of right-hyperbola by just using geometric reasoning (dilation, angles, etc) without using trigonometry, algebra, calculus, linear algebra, co-ordinate geometry. But doesn't prove the above mentioned conjecture along the way. The best I'm able to do is to just say that projection (dilation) of a conic section (in this case hyperbola) is still a conic section (which must be hyperbola because it was a broken curve before projection too). Can you guys come up with a more satisfying proof for above mentioned conjecture using only elementary geometric reasoning?

Edit: The direction of dilation can be either the axis passing through two Foci or the axis perpendicular to it.

Edit2: The "Dilation" in contrast to scaling, only streches or squishes all the lengths only along a particular direction. Think of it like streching a rubber sheet but only along one direction.

Answer 1

For a geometric intuition, you can use the fact that the hyperbola is a conic section. In particular, given a hyperbola in plane $\pi_1$, choose a right circular double cone whose axis is parallel to $\pi_1$ and that intersects $\pi_1$ in the given hyperbola.

Now if you dilate the entire space along the direction of the cone's axis, you still have a right circular double cone intersecting the dilated plane $\pi_1$ in a hyperbola.

If you linearly transform space by scaling all planes perpendicular to the cone's axis by the same amount, while keeping the distances between the planes fixed, you get another right circular double cone intersecting $\pi_1$ in a hyperbola. Meanwhile $\pi_1$ has been dilated in the direction perpendicular to the axis through the foci of the hyperbola. (This is equivalent to two dilations by the same amount, one in plane $\pi_1$ perpendicular to the axis through the foci, and one perpendicular to the plane. Note that the axis of the cone ends up at a different distance from the plane after the dilations.)

Answer 2

If $k$ is a dilatation factor, then if $P$ is a point on hyperbola with focuses $F$ and $G$ then we have $P'F' = kPF$ and $P'G'= kPG$ where $X'$ denotes image of $X$. So we have $$P'F'+P'G' = k(PF+PG) = constant$$ and thus $P'$ lies on (fixed) hyperbola with focuse $F'$ and $G'$.

Clearly, simillary is valid for any conic: image of a conic is again the same tipe of conic.

Edit: You can assume axsis of conic is $x$ axsis and $(0,0)$ is the center of conic. In that case you must replace $(x,y)$ by $(kx,ly)$ for some sutable fixed $k,l$.

Answer 3

Let us take hyperbola in a convenient form:

$${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2 ={\frac{x^2}{a^2}}-{\frac{y^2}{a^2(\epsilon^2-1)}}=1$$

Compare

$${\frac{x^2}{a^2}}-{\frac{y^2}{a^2(\epsilon_1^2-1)}}=1$$

with another hyperbola having dilations $(p,q) $ along $(x,y)$ directions

$${\frac{(xp)^2}{a^2}}-{\frac{(yp)^2}{a^2(\epsilon_2^2-1)}}=1$$

It is found to be still a hyperbola because its representation is in the same form. (Note how the constants are disposed..posiytive quantities in denominators below $(x^2,y^2)$.

Simplifying we get a relation between unstretched/stretched hyperbolas:

$$\frac{\epsilon_2^2-1}{\epsilon_1^2-1}=\frac{p^2}{q^2}$$

As a simple example let us start with the right-hyperbola aka the rectangular hyperbola:

$$\epsilon_1= \sqrt{2}; p=1.2,\,q=1.6,\,\frac{p}{q}=\frac 34\;$$

then the new hyperbola becomes a less eccentric

$$ \epsilon_2= 1.25 $$

which shows that the right-hyperbola can be dilated/squeezed to make it to become any other hyperbola in this way.

Answer 4

Let me add a proof that the image of a hyperbola under a generic dilation is another hyperbola (and an analogous proof can be devised for an ellipse or a parabola).

Suppose we are given any hyperbola of center $O$ (blue in figure below) and find its transformed (red) under a dilation having the direction of semidiameter $OA$ and ratio $k$. If $OB$ is the semi-diameter conjugated to $OA$ and $P$ is a generic point on the hyperbola, then we know from Apollonius' equation that

$$ {PH^2\over OA^2}-{OH^2\over OB^2}=1, $$ where $PH$ is parallel to $OA$ and $H$ lies on line $OB$.

Let then $P'$, $A'$, $B'$, $H'$, $O$ be the images of $P$, $A$, $B$, $H$, $O$ under the dilation. As $P'H'=k PH$ and $A'O=kAO$, then $P'H'/OA'=PH/OA$; amd by the intercept theorem we also have $OH'/OB'=OH/OB$. Hence:

$$ {P'H'^2\over OA'^2}-{OH'^2\over OB'^2}=1 $$

and it follows that $P'$ lies on hyperbola having $OA'$ and $OB'$ as conjugate semi-axes.

This proof doesn't work if the dilation has the direction of an asymptote: in that case one could use the asymptotic equation to give an analogous proof.