[RESOLVED] Parabolas & Directrix/Focus
Using the formula y=ax2 i can use a to find c, the distance to the focus and the directrix from the vertex, using the forumula a=1/4c
however i am trying to do this in code, so i need to isolate C on one side, and when i work it out i get c=1/4a, and that doesn't seem right...
Can some one enlighten me on this?
Re: Parabolas & Directrix/Focus
Resolved...
I worked it out, and they came out the same.
Assuming a=1/8:
using a=1/4c
1/8=1/4c
8=4c (reciprocal)
8/4 = 4c/4 (divide by 4, each side)
2 = c
using c=1/4a:
c=1/(4(1/8))
c=1/(1/2)
c=1 x 2 (multiply by reciprocal)
c=2
c = c
:)
Re: [RESOLVED] Parabolas & Directrix/Focus
Yup, hyperbolas like you've given (equations of the form x=q/y) are their own inverses. Not many functions have that property, so it's understandable that your result seemed strange.