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I'm aware of finding asymptotes of hyperbola using oblique asymptote but there is another way : put the standard equation of hyperbola equal zero (i.e. $ x^2/a^2 - y^2/b^2 = 0$) .

Why this method is right ? How we can explain it probably geometrically ?

S.H.W
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3 Answers3

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Algebraically, of course, you can simply factor the equation into $\left(\frac xa+\frac yb\right)\left(\frac xa-\frac yb\right)=0$ to see that it’s a pair of intersecting lines that are in fact the asymptotes of the hyperbola.

In two dimensions, the equation ${x^2\over a^2}-{y^2\over b^2}=z$ describes a family of hyperbolas parameterized by $z$ that all have the same asymptotes. You can think of this as taking horizontal slices through the three-dimensional hyperbolic paraboloid (saddle surface) given by that equation.

enter image description here When $z=0$, you’re taking a slice through the origin, which is the saddle point of this surface, and you end up with a pair of intersecting lines—the degenerate member of this family of conics.

This works for any hyperbola, in fact. If you rearrange its equation into the form $A(x-x_0)^2+2B(x-x_0)(y-y_0)+C(y-y_0)^2+F=0$, then by setting $F$ to $0$ you get an equation of its asymptotes.

amd
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One explanation would be this: as is well-known, the asymptotes of a curbe ‘meet the curve at infinity’. So consider the projective completion of the affine plane. A point in the projective plane is defined by its projective coordinates $[X:Y:T]$, not all $0$, defined up to a non-zero factor.

If $T\ne 0$, the point corresponds to a point $(x,y)$ in the affine space, such that $$x=\frac XT,\quad y=\frac YT.$$

If $T=0$, the point $[X: Y:0]$ is called the point at infinity in the direction $(X,Y)$.

Now to the affine hyperbola with equation $\;\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is associated its projective completion, which is defined by homogeneised equation: $$\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=T^2,$$ and to find the points at infinity of this projective hyperbola, we just set $T=0$, so that the points at infinity satisfy the equation $$\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=0.$$

Bernard
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  • Sorry but I'm not familiar with affine plane and projective coordinates . Can you explain more ? – S.H.W Jan 13 '18 at 16:31
  • The idea is this: on a line through the origin in the affine plane, with equation $y=ax$ you always can set $x=\frac XT,;y=\frac YT=a\frac XT$ for some $X, Y, T$, provided that $T\ne 0$. $X,Y,T$ are defined up to a non-zero factor. Observe that for fixed values of $X, Y$, the corresponding values of $x$ and $y$ get bigger and bigger when $T$ tends to $0$ – actually they tend to $infty$. That's why we say the pont with projective coordinates $[X:aX:0]$, which is the same as $[1:a:0]$ is the point at $\infty$ on the line $y=ax$, and the line with equation $T=0$ is the line at $\infty$. – Bernard Jan 13 '18 at 16:53
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Setting $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ and we consider the Case $$y=b\sqrt{\frac{x^2}{a^2}-1}$$ and we assume $$y=mx+n$$ is an equation for the asymptotes. then we have$$\lim_{x\to \infty}\frac{\left(\frac{b^2}{a^2}-m^2\right)x^2+2mnx-(b^2+n^2)}{b\sqrt{\frac{x^2}{a^2}-1}+mx+n}$$ this is equal to $$\lim_{x ßto 0}\frac{\left(\frac{b^2}{a^2}-m^2\right)x^2-2mnx-(b^2+n^2)}{\left(\frac{b}{a}+m\right)x}$$ and we get from here $$m=\frac{b}{a},n=0$$ so we obtain $$y=\frac{b}{a}x$$as one asymptote. The other one is $$y=-\frac{b}{a}x$$

Dr. Sonnhard Graubner
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