Asymptotes of hyperbola by considering x tends to infinity

Question

I saw this derivation of equations of the asymptotes of hyperbola and it goes like this...

For a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, rearranging the terms we get

$y=\pm\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$

So as $x\rightarrow\infty$, $\frac{a^2}{x^2}\rightarrow 0$ and hence $y\rightarrow\pm\frac{b}{a}x$.

Therefore the equations of the asymptotes are $y=\pm\frac{b}{a}x$.

I am not very sure if this derivation is correct. Would appreciate if you can share your opinion. Like if there is any step above that is not true in general.

Answer 1

Note that $$\frac{y}{\pm \frac{bx}{a}}=\sqrt{1-\frac{a^2}{x^2}},$$ and $\text{RHS}\to 1$ as $x\to\pm\infty.$ It then follows that $\text{LHS}$ is also $1$ at infinity. Thus, $$\frac{y}{\pm \frac{bx}{a}}\to 1,$$ which says that $y$ is asymptotic to the lines.

Answer 2

Here is what my teacher did. Solve the line $y=mx+c$ with the hyperbola by substitution. The resultant can be thought of as a quadratic in $x$. Since the line is an asymptote, it seemingly touches the hyperbola at infinity. The quadratic is: $$(b^2-a^2m^2)x^2 - 2a^2mcx -a^2c^2 - a^2b^2=0$$ Which has $\infty$ as both it's roots. Hence, the coefficient of both $x^2$ and $x$ should be zero, while constant term shouldn't. We get $3$ conditions as follows. $m=\pm \frac{a}{b}, 2a^2mc=0$(of which neither $a^2$ nor $m$ can be zero, leading to $c=0$), and $a^2b^2$ is not zero(as $c=0$) This leads to the equation of the asymptotes being $y=\pm \frac{a}{b} x$

Edit: This is to answer Allawonder's doubt. Assume that $\infty$ is a double root of the quadratic $ax^2+bx+c=0$. Then $0$ ought be a double root of the quadratic $cx^2+bx+a=0$(via transformation of equation). Hence, both $b$ and $a$ ought to be $0$ but $c$ can't be. P.S. Some people may have a hard time accepting this(I did too) but I got no choice except to trust my teacher.

Answer 3

Your derivation can be made correct by changing the final step. Consider your hyperbola: $$ y=\pm{b\over a}x\sqrt{1-{a^2\over x^2}} $$ and consider the couple of lines: $$ y=\pm{b\over a}x. $$ For a given $x$, the difference $\Delta y=y_{line}-y_{hyperbola}$ (of course you must subtract expressions with the same sign) is then $$ \Delta y=\pm{b\over a}x\left(1-\sqrt{1-{a^2\over x^2}}\right), $$ and you can check that $$\lim\limits_{x\to\pm\infty}\Delta y=0.$$

Answer 4

I think you are right to be suspicious of this method. Let's apply it to the hypberbola given by the equation

$$ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1 .$$

Solving for $y$ as in the method in the question, we get

$$ y = \pm x \sqrt{1 + \frac 4x - \frac 4{x^2}},$$

and as $x\to\infty,$ we find that $\sqrt{1 + \frac 4x - \frac 4{x^2}}\to 1,$ so this method derives the asymptotes $y = x$ and $y = -x.$ But the actual asymptotes are $y = 2 + x$ and $y = 2-x.$

The pitfall in this method can be seen by applying it to any straight line. For the equation $y = mx + b,$ We factor $x$ out of the right side to obtain

$$ y = x \left(m + \frac bx\right), $$

and then $\left(m + \frac bx\right)\to m$ as $x\to\infty,$ so the method yields $y = mx.$

What the method is actually finding is the directions from the origin to the points at infinity on the curve, which gives the slopes of the asymptotes but not the $x$- or $y$-intercepts. It will coincidentally give the correct result when the asymptote happens to pass through the origin, as we can predict will happen with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ due to symmetry, but that's an extra fact that has to be shown and it works only in that special case.

However, the method does find the slope of each asymptote. We can then find the $y$-intercept by taking the difference between the curve and a line through the origin with the same slope as the asymptote. Taking the hyperbola $ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1$ again, taking the solution $y = \sqrt{x^2 + 4x -4}$ and comparing it with the line $y = x,$ we find that $$ \lim_{x\to\infty} (y_\mathrm{\,hyperbola} - y_\mathrm{\,line}) = \lim_{x\to\infty} \sqrt{x^2 + 4x -4} - x = 2.$$

Hence the difference between the curve $y = \sqrt{x^2 + 4x -4}$ and the line $y = 2 + x$ as $x\to\infty$ is zero. This correctly predicts that $y = 2 + x$ is an asymptote.

Aretino's answer shows how this method is correctly applied to the hyperbola in the question.