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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

Blue
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Archis Welankar
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  • Actually, if your point is “inside” the hyperbola, there is no tangent drawable to it. But does it surprise you that there are only two tangents to a circle from a point outside it? If not, you might be interested to know that circle and hyperbola are (projectively) equivalent in a way that preserves tangency and the possibility of drawing tangents. – Lubin Aug 10 '16 at 14:12
  • From outside the hyperbola I know it seems like four tangents can be drawn but if you extend the tangents on both sides you will find 2 out of the 4 tangents you imagined actually overlap. –  Aug 13 '16 at 14:38
  • Ya like the x formed has two tangents same – Archis Welankar Aug 13 '16 at 15:22
  • @Lubin can you justify that there can be drawn no tangent from centre of hyperbola to hyperbola – imposter Jun 10 '21 at 11:18
  • Projectively @Anusha, there are tangents to the hyperbola from the center: they’re the asymptotes. Notice that if $P$ is one one of the asymptotes but not the other, there’s only one tangent in the finite plane. The other tangent is the asymptote that $P$ lies on. – Lubin Jun 10 '21 at 15:52
  • @Lubin yup thank you – imposter Jun 11 '21 at 15:45

2 Answers2

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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''.

Emilio Novati
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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.

David Quinn
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