Conics Problem Part 2: Formulas and Solutions | Homework Statement

  • Thread starter Thread starter temaire
  • Start date Start date
  • Tags Tags
    Conics
AI Thread Summary
The discussion focuses on solving a conics problem related to hyperbolas. The participant confirms that their solution for part (i) is correct but questions their approach for part (ii), specifically in calculating the value of b. They clarify that the formula b^2 = c^2 - a^2 applies, with c being the distance to the foci and a being the distance from the center to the vertex. Upon realizing a misunderstanding regarding the hyperbola's equation form, they correct their values for a and b. The final conclusion is that b is correctly identified as 10, aligning with the problem's requirements.
temaire
Messages
275
Reaction score
0

Homework Statement


This is the second page.
http://img116.imageshack.us/img116/7519/arch2iq9.jpg​
[/URL]


Homework Equations


Formulas on picture above.


The Attempt at a Solution


I'am just wondering if everything looks fine.
 
Last edited by a moderator:
Physics news on Phys.org
part (i) is correct.
For part (ii), I don't think that's how you would solve for b. There's a formula to solve for b. b^2=c^2-a^2 where c is your focii and a is the distance from your center (0,30) to the vertex of the hyperbola. a=10, c=30, then b=?
 
Whoops, I realized that the problem is using a different form of the vertical hyperbola instead of \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1, so in this case I switched a and b around.

In this problem, where b=10(the distance from center to vertex of hyperbola) is correct.
 
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...
Back
Top