Are hyperbolas of the same eccentricity shaped different?

Question

It seemed like that shouldn't be the case, because as I understand it, figures retain their shape when each coordinate is scaled by the same factor. So if I'm keeping the eccentricity of the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)}=1$$ constant and multiplying $a$ by, say, 3, it seems like both coordinates of each point would be similarly scaled by 3, and that the hyperbola would, as above, retain its shape.

But this doesn't seem apparent when I graphed it (I used q for the eccentricity because Desmos thinks e is 2.71...). When you change a from the 'hyperbola' folder, you can see the hyperbola squeeze at the centre.

I know that, as with parabolas, it may not be immediately apparent that multiple figures are shaped the same. But if hyperbolas of different eccentricities are shaped similar - and the bit about uniform scaling not changing the shape is correct - why does it look like this?

And if either or both points are wrong, why so?

Answer 1

Here is the hyperbola with $a=3.2, e=1.34$
Here is the hyperbola with $a=1.6, e=1.34$
You say they are not similar because it is "squeezed in the center".

If we take the center part of the $1.6$ picture, and magnify by factor $2$ we get
And now it definitely does look like the $3.2$ picture, right? If you superimpose them, they match exactly.

Answer 2

Yes, all hyperbolas with the same given eccentricity are similar. Your reasoning is perfectly sound, but perhaps your eyes deceive you.

I had a look at your Desmos construction, and it is consistent with what we should expect to see. As parameter $a$ is changed, it acts as scale factor for a dilation. The curve crowds closer to the center as $a$ approaches zero. Well, of course it does. We should expect that behavior in a dilation.

The asymptotes are invariant. We should expect that too. They are lines through the center of dilation.