Focus of a Parabola
12-27-2018, 04:57 PM
Post: #1
 MullenJohn Junior Member Posts: 43 Joined: Sep 2017
Focus of a Parabola
Is there any way to find the focus of a parabola (e.g. 2x^2 -x -1) using the HP Prime. I cannot find a way to do it. If you have found a way please tell me the steps. Cheers!
12-27-2018, 06:13 PM
Post: #2
 toshk Member Posts: 195 Joined: Feb 2015
RE: Focus of a Parabola
(12-27-2018 04:57 PM)MullenJohn Wrote:  Is there any way to find the focus of a parabola (e.g. 2x^2 -x -1) using the HP Prime. I cannot find a way to do it. If you have found a way please tell me the steps. Cheers!

http://www.hpmuseum.org/forum/thread-10605.html
Conics(y-(2*x^2-x-1))
12-27-2018, 06:34 PM
Post: #3
 MullenJohn Junior Member Posts: 43 Joined: Sep 2017
RE: Focus of a Parabola
Thank You Toshk.
I was looking for a built-in HP Prime function like something under Catlg or CAS, et cetera. Is there anything like finding the focus with a built-in HP Prime function? Cheers!
12-27-2018, 06:37 PM
Post: #4
 toshk Member Posts: 195 Joined: Feb 2015
RE: Focus of a Parabola
(12-27-2018 06:34 PM)MullenJohn Wrote:  Thank You Toshk.
I was looking for a built-in HP Prime function like something under Catlg or CAS, et cetera. Is there anything like finding the focus with a built-in HP Prime function? Cheers!

reduced_conic(y-(2*x^2-x-1))
12-28-2018, 02:37 PM
Post: #5
 MullenJohn Junior Member Posts: 43 Joined: Sep 2017
RE: Focus of a Parabola
Thank you Toshk. I was unaware of the function reduced_conic().

I was unable to determine from the results of reduced_conic(y-(2x^2+3x+1)) which item would tell me the focus (-3/4, 3/8) or the focal length (p = 1/8).

Which part of the result states the focus coordinates or the focal length p = 1/4a?

Cheers!
 « Next Oldest | Next Newest »

User(s) browsing this thread: 1 Guest(s)