Why only 2 tangents can be drawn to a hyperbola?

Question

Today we were introduced to what's an hyperbola its properties and hence equation of tangents, normal, pole, polar.

I was stunned when my professor said that we can draw only two tangents to a hyperbola (whether same branch or different). So my confusion begins here.

Now equation of tangent in parametric form is $\frac{x\sec(\theta)}{a}-\frac{y\tan(\theta)}{b}=1$ . Reducing it with double angle formulae, I got quadratic in $\tan(\theta/2)$ which means two tangents can be drawn. But if you just see a hyperbola without much details, a beginner like me would glance and say $4$ tangents can be drawn (with common logic).

What condition is it that justifies that, at maximum, only two tangents can be drawn except the way on which we proved in class. I am probably looking for a geometric clarification.

Thanks

Answer 1

If I well understand you want to prove that we can have at most two tangent lines to a conic passing through a common point. This is a simple consequence of the fact that the equation of a conic is second degree.

Let $f^{(2)}(x,y)=0$ be the equation of the conic and $P=(x_P,y_p)$ a point. The stright lines between $P$ are represented by the equation $y-y_P=m(x-x_P)$ and the common points of any such lines with the conic are the solutions of the system: $$ \begin{cases} f^{(2)}(x,y)=0\\ y-y_P=m(x-x_P) \end{cases} $$

Substituting $y$ from the equation of the lines in the first equation we find an equation $R^{(2)}(x,m)=0$ that is of second degree in $x$ and contains the parameter $m$. The line is tangent to the conic iff this equation has only one real solution, i.e. if the discriminant $\Delta_R^{(2)}(m)=0$.

Since the discriminant is a quadratic function of the coefficients, this is a second degree equation in $m$ and we have three possibilities:

1)the equation: $\Delta_R^{(2)}(m)=0$ has no real solutions. This menas that we have no tangent lines form $P$. We say that $P$ is an ''internal'' point of the conic.

2) The equation has one (double) real solution: there is only one tangent line and the point $P$ is a point of the conic.

3) We have two distinct real solutions of the equation, so we have two tangent lines and the point $P$ is ''external''.

Answer 2

If we take the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ suppose that $y=mx+c$ is a tangent.

Solving simultaneously, eliminating $y$, leads to the quadtatic equation in $x$:

$$(b^2-a^2m^2)x^2-2a^2cmx-(a^2c^2+a^2b^2)=0$$

Setting the discriminant equal to zero and simplifying leads to the condition:$$c^2+b^2=a^2m^2$$

Therefore for a given $(x,y)$ through which the tangent must pass, we are left with a simultaneous pair of equations, quadratic in $m$ and $c$, which will lead to at most two solutions, and therefore at most two tangents.