What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$? 
What is the angle between the asymptotes of this hyperbola?
  $$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$

I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?
 A: Solve the quadratic equation for $y$:
$$y=-x\sqrt{7}-\sqrt{12x^2-2x+1},y=-x\sqrt{7}+\sqrt{12x^2-2x+1}$$
Let the assymptotes be $y=a_1x+b_1$ and $y=a_2x+b_2$. Then:
$$\lim_{x\to\infty} \frac{-x\sqrt{7}-\sqrt{12x^2-2x+1}}{a_1x+b_1}\stackrel{LR}{=}\lim_{x\to\infty} \frac{-\sqrt{7}-\frac{12x-1}{\sqrt{12x^2-2x+1}}}{a_1}= \lim_{x\to\infty} \frac{-\sqrt7-2\sqrt3}{a_1}=1\Rightarrow \\
a_1=-\sqrt7-2\sqrt3$$
Similarly, $a_2=-\sqrt7+2\sqrt3$. The angle between the lines with such slopes is:
$$\tan \phi=\left|\frac{a_1-a_2}{1+a_1a_2} \right|=\sqrt3 \Rightarrow \phi =60^\circ.$$
A: Let the slope of an asymptote be $m$.
The quadratic terms of the conic yield
$$5-2\sqrt7m-m^2=0.$$
Hence by Vieta, the product of the slopes is $-5$, and their sum $-2\sqrt7$. The difference is $\pm\sqrt{2^2\cdot7-4\cdot(-5)}=\pm4\sqrt 3$.
Hence the tangent of the angle,
$$\tan(\phi)=\frac{m_+-m_-}{1+m_+m_-}=\pm\frac{4\sqrt3}{1-5}=\pm\sqrt3.$$
A: Hint:
Compare $$(m_1x-y+c_1)(m_2x-y+c_2)=m_1m_2x^2-2xy(m_1+m_2)+y^2+\cdots$$
with $$-5x^2+2\sqrt7xy+y^2+\cdots=0$$
$$\dfrac{m_1m_2}{-5}=\dfrac{-2(m_1+m_2)}{2\sqrt7}=\dfrac11$$
Use
$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$$
We need $$\arctan\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$
A: Rearranging lab's answer ...  
Omit the lower-degree terms of $5x^2-2\sqrt 7 xy-y^2-2x+1=0$ to get $5x^2-2\sqrt 7 xy-y^2=0$.  This is the equation of two lines which are parallel to your asymptotes.  So find the angle between these two lines.  
Factor:
$$
5x^2-2\sqrt 7 xy-y^2 =
-\frac{\left( 2\,y\sqrt {3}-\sqrt {7}y+5\,x \right)  \left( 2\,y\sqrt 
{3}+\sqrt {7}y-5\,x \right) }{5}
$$ 
Solving and rationalizing, the lines are
$$
y=(2\sqrt{3}-\sqrt{7})x,\quad y=(-2\sqrt{3}-\sqrt{7})x
$$
Use the addition formula for tangents as lab suggests.
