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 $$ y=\pm (b/a)x $$