Reason for setting coefficients of $x$ and $x^2$ equal to zero when roots are infinite. While studying coordinate geometry, I came across a topic to find the equation of asymptote of a standard hyperbola, $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$ in that topic, the equation of asymptote was assumed to be $y = mx + c$, and them after some work (forming a quadratic equation having roots that gave $x$ coordinates of the point of intersection of line and hyperbola), the coefficient of $x$ squared and $x$ was set to be zero due to the fact that asymptote never intersects the curve. (So the roots of quadratic that is , the $x$ coordinates of point of intersection will tend to infinity).

I want to know the REASON behind this analogy, although I'm guessing since the sum of roots is given to be the negative of coefficient of $x$ DIVIDED by coefficient of $x$ squared (by Veita's formula), since the sum of roots tends to infinity, the denominator that is, the coefficient of x squared must tend to zero, is my assumed analogy correct? But then, why the coefficient of $x$ is set to zero? 

Any help will be greatly appreciated. 

EDIT:

Here's the photo of the page of book in which equation of asymptote was derived. (https://i.stack.imgur.com/Via9z.jpg)


  the straight line $y = mx + c \tag 1$ meets the hyperbola in points, whose abscissae are given by the equation $\qquad x^2(b^2 - a^2m^2) - 2a^2 mcx - a^2(c^2+b^2) = 0 \tag 2$ If the straight line be an asymptote, both the roots of $(2)$ must be infinite.  Hence, the coefficients of $x^2$ and $x$ in it must both be zero.
What is going to happen if the coefficients of $x$ and $x^2$ are not zero?

$$\qquad x^2(b^2 - a^2m^2) - 2a^2 mcx - a^2(c^2+b^2) = 0 \tag 2$$

As $x\to \infty$ the right hand side stays at $0$ while the left hand side approaches either infinity or negative infinity depending on the other parameters.

Another way to see the asymptotes is to solve$$ \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1$$  for $y$

Note that $$ y=\pm (b/a)x\sqrt {1-(a/x)^2}$$ which is asymptotic to  $$ y=\pm (b/a)x $$
If you now know why the coefficient of $x^2$ is zero here's an explanation for why coefficient of $x$ is zero

asymptotes are tangents to  the hyperbolas at infinity and therefore has discriminant  equal to zero
i.e. $b^2=4ac$     in equation $ax^2 + bx +c$
      $(a =0)$
     now to satisfy this condition $b=0$