The equation of an asymptote can be either $$y=\pm\frac{b}{a}\sqrt{ (x-h)} + v.$$ The $v$ tends to be ignored as trivial, as the $x$ value tends to infinity, which implies that the approximate asymptote shall always be less than the actual one. However, if we did graph the hyperbola accurately, how would the asymptotes vary?
Asked
Active
Viewed 86 times
0
-
1Please use MathJax to typeset the equations nicely. That way, we'll know precisely what you mean. – Matti P. Apr 15 '19 at 10:15
-
1What equation of the hyperbola are you using? With the standard equation $$\frac{(x-h)^2}{a^2}-\frac{(y-v)^2}{b^2}=1$$ the asymptotes are $$y=\pm\frac{b}{a}(x-h)+v$$ It's true that, as $x$ grows without bound, $v$ becomes negligible, but that's true of any line equation $y=mx+v$. That doesn't mean you can drop the $v$ from the line equation itself; the value remains significant at smaller scales. – Blue Apr 15 '19 at 13:43
-
Yes, but the question I wish to ask is how one graph the asymptote, if we did not let v become negligible. – LÜHECCHEgon Apr 15 '19 at 19:50
-
The asymptotes intersect at the hyperbola’s center, which you can read from the equation, while their slopes are $\pm b/a$. – amd Apr 16 '19 at 05:36
2 Answers
0
If the equation of the hyperbola is $$\frac{(x-h)^2}{a^2}-\frac{(y-v)^2}{b^2}=1,$$ then a common equation of asymptotes is $$\frac{(x-h)^2}{a^2}-\frac{(y-v)^2}{b^2}=0.$$ From here we easily deduce both equations of asymptotes, as written in the comment of Blue.
user376343
- 8,311
0
You are perhaps meaning the $\delta y$ 's but not $v$'s themselves. Also there is no square root in the pair of straight lines that represent asymptotes in the shifted situation.
$$ y=\pm\frac{b}{a} (x-h) + v $$
where the original asymptotes are
$$ y=\pm\frac{b}{a} x $$
In a linear transformation of $(x,y)$ variables the entire blue graph of the hyperbola $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
is rigidly translated/displaced by addition of the displacement vector $(h,v)$ the vector as red graph.
$$ \frac{(x-h)^2}{a^2}-\frac{(y-v)^2}{b^2}=1$$
Neither $h$ nor $v$ tends to get neglected as trivial.
The vertical ( and horizontal ) difference of the hyper and its asymptote at corresponding points (blue to red) remains always the same, whatever may be its magnitude.They are marked at two places.
It is as if you took two identical overhead transparent projection plastic sheets and parallelly slided one sheet axis relative to the other axis through the displacement vector $(h,v)$ at origin as well as at other points without any rotation of axes. The vertical blue and red segments shown are equal in length.
Narasimham
- 40,495