How would you find the equations of the asymptotes

  • Thread starter Thread starter chaoseverlasting
  • Start date Start date
AI Thread Summary
The discussion focuses on finding the equations of asymptotes for hyperbolas, specifically through the equations L_1L_2 + λ = 0 and S + λ = 0, which represent linear functions and straight lines, respectively. There is a query about whether the first two equations are equivalent and how to derive asymptotes from a given curve equation. The third equation, S + 2λ = 0, is identified as representing the conjugate hyperbola if S + λ = 0 denotes the original hyperbola. Clarification is sought on whether the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\) indeed represents the conjugate hyperbola, confirming that it does. The conversation highlights the complexities involved in understanding these mathematical relationships.
chaoseverlasting
Messages
1,050
Reaction score
3
L_1L_2 +\lambda =0
S+\lambda =0 (Such that D=0)
S+2\lambda =0

In case of a hyperbola, S is the pair of straight lines representing the asymptotes and \lambda is any parameter.

My question is, are the first two equations the same? How would you find the equations of the asymptotes if you were given the equation of the curve.

The third equation is the conjugate hyperbola if S+\lambda =0 represents the original hyperbola. Is there any other way to find the conjugate hyperbola?

If \frac{x^2}{a^2} - \frac{y^2}{b^2} =1 is the equation of the original hyperbola, then does the equation \frac{x^2}{a^2} - \frac{y^2}{b^2} =-1 represent the conjugate hyperbola?
 
Physics news on Phys.org
I can make very little sense of this. What are L1 and L2? Are they linear functions? What is S?
 
L_1 and L_2 are linear functions. S represents a pair of straight lines.
 
Yes, to your very last part. I found the rest to be confusing.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top