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
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.
The hyperbola ${x^2 \over a^2} - {y^2 \over b^2} = 1$ has the asymptotes $y = \pm {b \over a} x$. Hence $\tan {\alpha \over 2} = {b \over a}$ or ${a \over b}$ according to your choice of the angle between the asymptotes. Then just use the identity $\sec^2 x = 1 + \tan^2 x$.
If you want $\sec {\alpha \over 2}$ to be in terms of $e$, then you'll have to choose $\alpha$ to be that angle between the asymptotes which contains the $x$-axis. Then, you'll get
$$ \sec {\alpha\over 2} = e. $$
For a method to find the equation of the pair of asymptotes to a hyperbola:
Let the equation of the hyperbola be:
$$ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0. $$
Note that the equation of a hyperbola and its pair of asymptotes differ only in their constant terms. So, we write the equation of the pair of asymptotes as:
$$ ax^2 + 2hxy + by^2 + 2gx + 2fy + \lambda = 0 $$
If the above equation is the equation of a pair of straight lines then $h^2 ≠ ab$ and,
$$ ab\lambda + 2fgh - af^2 - bg^2 - \lambda h^2 = 0. $$
You get the value of $\lambda$ from here.
