I saw this solution to finding the equations of the asymptotes of the hyperbola of form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. To find the asymptotes, you let the right hand side equal 0, then rearrange to get your equation. Like this: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0$, then $\frac{x^2}{a^2} = \frac{y^2}{b^2}$, then $\pm\frac xa = \frac yb$, and thus $ y = \pm \frac bax$. I do not understand why you set it equal to $0$, though. I get that the asymptotes shouldn't be part of the graph, so we set it to zero to find the points not on the hyperbola, but why set it to 0?
Asked
Active
Viewed 700 times
1
-
Esssentially a duplicate of https://math.stackexchange.com/questions/3698030/where-does-the-equation-of-asymptotes-of-a-hyperbola-come-from , so also see the answers there – Toffomat Jan 11 '21 at 08:24
2 Answers
3
An asymptote denotes a tendency, when a variable goes to infinity.
When $x$ or $y$ tends to infinity the finite terms contribute only insignificantly . Dividing by $y^2$
$$ \frac{x^2}{a^2 y^2} - \frac{1}{b^2} = \frac{1}{y^2} $$
In order to trace to what curve the graph tends to, we need to a priori set the small finite fraction ... including the increasing denominator ... to zero.
Narasimham
- 42,260
0
Hint:
See $\#313$ of The elements of coordinate geometry
From the $\#324$ of the same, the equation to the asymptotes only differs from any hyperbola by a constant.
So, here the equation of the pair of asymptotes here will be $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=K$$ where $K$ is any arbitrary constants
Now use What is condition for second degree equation to represent a pair of straight lines?
$\begin{pmatrix} \dfrac1{a^2} & 0&0\\0&\dfrac1{b^2}&0\\0&0&-K\end{pmatrix}=0\implies \dfrac{-K}{a^2b^2}=0$
lab bhattacharjee
- 279,016