Why are there only two tangents to a hyperbola from a point, instead of four?

Question

Why are only two tangents possible to a hyperbola from a point? If we treat the hyperbola as two individual parabolas, then a point should be able to create two tangents through it for both of them, hence a total of 4.

Answer 1

The diagram shows a point $P$ between the arms of a hyperbola. It also shows how the rays emanating from $P$ fall into four regions (denoted "North", "South", "East", "West") bounded by $\overrightarrow{PA}$, $\overrightarrow{PB}$, $\overrightarrow{PC}$, $\overrightarrow{PD}$:

Rays in the North and South regions never hit the hyperbola at all. Rays in the East and West regions cut the hyperbola.

The boundary rays are special.

• $\overrightarrow{PA}$ and $\overrightarrow{PB}$ are tangent to the hyperbola (at "finite" points).

• $\overrightarrow{PC}$ and $\overrightarrow{PD}$ are parallel to the (invisible) asymptotes of the hyperbola. (It's often helpful to think of these as being tangent to the hyperbola at the "point at infinity", but that's not the sense in which you're using tangency.)

• (It's also good to know that the ray opposite $\overrightarrow{PC}$ separates the "Western" rays into two sub-regions: those that meet the hyperbola once, and those that meet it twice; likewise, the ray opposite $\overrightarrow{PD}$ sub-divides the "Eastern" rays.)

Moving $P$ can alter the nature of the boundaries a bit; for instance, if $P$ lies directly between the vertices (but, say, "west" of center), then $\overrightarrow{PA}$ and $\overrightarrow{PD}$ become the tangents, while $\overrightarrow{PB}$ and $\overrightarrow{PC}$ are asymptote-parallels. But the overall idea holds here, and apart from the case in which $P$ coincides with the hyperbola's center, there are always two tangents and two asymptote-parallels.

Anyway, the fact that a hyperbola has asymptotes (and thus region-bounding rays parallel to those asymptotes) distinguishes the geometry here from that of "two parabolas", which would have no asymptotes (and thus no corresponding region-bounding rays).

Answer 2

(see figure) @wojowu @Blue @Shreyash Chaudhari

Here is a mathematical explanation, using projective geometry.

Let us consider the case of an ellipse : from a given point outside an ellipse, you can draw two tangents (only).

If you have done projective geometry, you know that one can transform any conic section into any other one, in particular, one can map a hyperbola onto an ellipse, and vice versa.

As contact points and their tangents are preserved in such a transform, we are done.

With an example, it will be better; consider projective transform (P):

$$\begin{cases}X=\dfrac{2x}{x+y-1}\\Y=\dfrac{y}{x+y-1}\end{cases}$$

(P) "sends" the ellipse onto the hyperbola, sends $A(x,y)=(-1, 1)$ onto $B(X,Y)=(2,-1)$ and maps the tangents issued from $A$ to the ellipse onto the tangents issued from $B$ to the hyperbola.

I give below the Matlab program that I have done for this picture.

clear all;close all;hold on; grid on;

set(gcf,'color','w');

set(0,'defaulttextfontsize',16);

set(0,'defaultlinelinewidth',2);

axis([-2,3,-2,2]);

Tx=@(x,y)(2*x./(x+y-1)); % proj. transf. equ.

Ty=@(x,y)(y./(x+y-1)); % contd

t=0:0.01:2*pi;

x=1+2*cos(2*t-pi/4); % ellipse's param. equ.

y=sin(2*t); % contd

plot(x,y);

plot(Tx(x,y),Ty(x,y),'r')

x=[-1,-1,2.4];y=[-0.7,1,1];

plot(x,y,'-.b'); % draw tangents to ellipse issued from A

plot(Tx(x,y),Ty(x,y),'-.r'); % draw tangents to hyperbola issued from B

scatter([-1,2],[1,-1],30,'filled','k')

text(-1.3,1.3,'A');text(2.1,-1.3,'B');