Find eccentricity and directrix
r=5/(-1+2cosq)

Question

Find eccentricity and directrix
r=5/(-1+2cosq)

experimentx · Answer

The standard equation of the conic is given by
$$ \frac{l}{r} = 1 + e\cos\theta$$

experimentx · Answer

eccentricity = 2

anonymous · Answer

directrix is -5/2 or 5/2 ??

experimentx · Answer

are you trying to get directrix in polar form or in xy form??

anonymous · Answer

directrix is x=-5/2 or 5/2 ?

experimentx · Answer

must be 5/2 http://www.wolframalpha.com/input/?i=polar+plot+r+%3D5%2F%28-1%2B2cos heta%29

anonymous · Answer

THX

experimentx · Answer

wait, i don't think it should be 5/2

experimentx · Answer

It should be a little beyond 3 http://www.wolframalpha.com/input/?i=polar+plot+r+%3D5%2F%281%2B2cos+theta%29%2C+r+%3D5%2F%28-1%2B2cos+theta%29

anonymous · Answer

i think ed = 5 then e=2
so d = 5/2

anonymous · Answer

center is 10/3

experimentx · Answer

NO ...
the equation of the directrix should be ... something like
$$ \frac{\sqrt{r^2 - (5/2)^2}}{r} = \cos \theta$$

or 

$$ \frac{(5/2)}{r} = \sin \theta$$

experimentx · Answer

yeah ... centre is 10/3

anonymous · Answer

i am using polar equation
r=ed/1-ecosq
d=directrix

experimentx · Answer

Ooops ... sorry!! i was wrong!!

anonymous · Answer

no nevermind thanks for your help anyway

experimentx · Answer

but this is hyperbola ... shouldn't the directrix pass though the centre??

anonymous · Answer

there are two directrix 5/2 and 25/6

anonymous · Answer

center is in between of two directrix

experimentx · Answer

I guess it is then ..

experimentx · Answer

I don't understand why this is working http://www.wolframalpha.com/input/?i=plolar+plot+sqrt%28r^2+-+%285%2F2%29^2%29%2Fr%3D+sin+theta+from+theta+%3D+-pi%2F6+to+theta%3Dpi%2F6

experimentx · Answer

http://www.wolframalpha.com/input/?i=plolar+plot+%285%2F2%29%2Fr%3D+cos+theta+from+theta+%3D+-pi%2F6+to+theta%3Dpi%2F6

anonymous · Answer

i found a video http://www.youtube.com/watch?v=zV6buIBpuJo