How we calculate the asymptotes of a general hyperbola?

Question

I was reading about the asymptotes of the following hyperbola:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \quad (1)$$ and the book said that the inclined asymptotes are $ y = \pm \frac{b}{a} x $ and the book mentioned that you can find them by setting the RHS of eqn.(1) equal to $0$ but I do not know why we should do this, Could anyone explain this for me please?

Also, I know that the correct way of finding the inclined asymptote (y = mx + c) for a function $f(x)$ is that if this line satisfies $$\lim_{x \to \pm \infty} [f(x) - (mx + c)] = 0.$$ Then it is an asymptote. but I do not know how we found that line in the case of the hyperbola mentioned above. could anyone explain this for me please?

Answer 1

DISCLAIMER: This is based on a thought and intuition I have on asymptotes, and not mentioned in any book I have read. So this might sound unappealing to some, but I am afraid I do not have any sources to confirm that this method is correct...I just don't find anything wrong in it.

This may not be the most elegant way to look at it, but if you consider any hyperbola: $$ax^2+by^2+2gx+2fy+2hxy+c=0 \ \ or \ \ \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$$

The asymptotes to it will pass through its centre.

What I say next might not be an actual thing, but just an intuition I have towards asymptotes: Asymptotes behave the same way as the hyperbola at infinity (following along the line of the asymptote), so if $H=0$ is the hyperbola and $A=0$ are equations of asymptotes, then I think as we approach infinity (let's call the point we are approaching here as P) along the asymptote, $A_{at P}=H_{at P}$ (Again just an idea or way of thinking, may not be accurate). So at such a point P, the terms in A and H that actually contribute to the value of A or H are the ones that have variables $x$ and $y$ in them as the are the ones blowing up to $\infty$.

So if the two are same at P, the coefficients of terms containing $x$ and $y$ will have to be same.

In other words, the constant term is the only thing that differs in the two equations A and H.

So if $A=0$ is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then we can write the equation of asymptotes as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\lambda \ (some \ constant)$

Also since the asymptotes pss through the centre of the hyperbola ($(0,0)$ in this case), put x=0 and y=0 in equation.

So $\lambda = 0$.

That is where I believe the transformation of 1 to 0 came from in your book.

Answer 2

HINT.-$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Making $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0\iff(\frac xa-\frac yb)(\frac xa+\frac yb)=0$.

so you do have $y=\pm\dfrac{bx}{a}\quad\quad (1)$ $$y=\pm b\sqrt{\frac{x^2}{a^2}-1}=\pm\frac{bx}{a}\sqrt{1-\frac{a^2}{x}}\quad\quad (2)$$

To what tends $(2)$ when $x\to \infty$?