analytic geometry
If $\alpha$ is the angle between the asymptotes of hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$ with eccentricity $e,$
then $\displaystyle \sec \frac{\alpha}{2}$ is
assuming $y=mx+c$ is the equation of hyperbola
asymptotes is the line which touches curves at infinity
$\displaystyle \frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2} = 1$
$\displaystyle (b^2-a^2m^2)x^2-2a^2mcx-(c^2+a^2b^2)=0$
wan,t be able to further after that, could some help me with this? Thanks
This is an euclidean solution to the parabolic case.
To adapt the proof to the hyperbolic case is left to the reader.
Lemma 1 (how to draw a tangent to a parabola). Let $A$ the projection of $P$ on the axis of the parabola and $B$ the symmetric of $A$ with respect to $V$. Then $PB$ is the tangent to the parabola at $P$.
In modern terms, this is just $\frac{d}{dx}x^2 = 2x$.
Lemma 2 (optical property of the parabola). If $C$ is the projection of $P$ on the directrix, the tangent at $P$ bisects the angle $\widehat{FPC}$.
Proof: Let $O$ be the intersection between the axis and the directrix. By Lemma 1 we have $PAF=COB$, so $PFBC$ is a parallelogram. Since $PF=PC$ by the definition of parabola, $PFBC$ is indeed a rhombus.
Corollary. Since $PF=PC$ and $PQ$ bisects $\widehat{FPC}$, $C$ and $F$ are symmetric with respect to $PQ$. It follows that $$ \widehat{PFQ}=\widehat{PCQ} $$ so $\widehat{PFQ}$ is a right angle.
In the hyperbolic/elliptic case Lemma 2 and the next Corollary have to be replaced with: the tangent at $P$ is the internal/external angle bisector of $\widehat{F_1 P F_2}$, hence $PCQF$ is a cyclic quadrilateral and $\widehat{PFQ}=\widehat{PCQ}$ holds.
(Note: this answer shows how to work it out using the supplied formula for the tangent, so I'll keep it up, but there's a much simpler graphical method here.)
First off, there's an typo in your formula for eccentricity. It should be $\sqrt{(a^2 + b^2)/a^2}$.
With that out of the way, the key point to figuring this out is to realize that both the eccentricity and the angle really only depend on the ratio $b/a$, as they both measure how "squashed" the hyperbola is. So our job is now figuring out how to massage the formulas so they only depend on $b/a$, and not on $a$ or $b$ individually. For the tangent of the angle
$$\dfrac{2ab}{a^2-b^2} = \dfrac{1/a^2}{1/a^2}\cdot \dfrac{2ab}{a^2-b^2} = \dfrac{2ab/a^2}{a^2/a^2-b^2/a^2} = \dfrac{2(b/a)}{1-(b/a)^2}. $$
You can do something similar for the eccentricity formula.
Can you take it from there? Me, once I've rewritten the formulas in terms of $b/a$, I might introduce a new variable $k= b/a$ just to make the subsequent manipulations easier.
Best Answer
HINT: The slope of the asymptotes is $\pm\tan\frac{\alpha}2$, from which you can find $\sec\frac{\alpha}2$ via the identity $1+\tan^2\theta=\sec^2\theta$. This slope can be computed by finding the limit of $y/x$ for points on the hyperbola as $x\to\infty$. (You can either solve the hyperbola’s equation for $y$ or use $x=a\cosh t$, $y=b\sinh t$ and let $t\to\infty$.) Once you have the answer in terms of $a$ and $b$, see if you can recognize it as some other property of the hyperbola.