# Foci

• Apr 23rd 2011, 03:34 PM
rai2003
Foci
r(\phi ) = 3/( 4+ 2sin(\phi ) for 0\leqslant \phi \leqslant 2\pi

1 .Name of the shape of the graph
2. One foci is at the origin
Find the other foci is at ( , )
The first question i think the shape of graph is Hyperpolar since a>1
• Apr 23rd 2011, 05:45 PM
topsquark
Quote:

Originally Posted by rai2003
r(\phi ) = 3/( 4+ 2sin(\phi ) for 0\leqslant \phi \leqslant 2\pi

1 .Name of the shape of the graph
2. One foci is at the origin
Find the other foci is at ( , )
The first question i think the shape of graph is Hyperpolar since a>1

One reason you might not be able to solve for the foci is that the graph is not a hyperbola, it is an ellipse.

-Dan
• Apr 23rd 2011, 07:32 PM
rai2003
find Foci
Quote:

Originally Posted by topsquark
One reason you might not be able to solve for the foci is that the graph is not a hyperbola, it is an ellipse.

-Dan

Can u tell me how i can find Foci since i know this is ellipse. Thanks
• Apr 24th 2011, 05:32 AM
topsquark
Quote:

Originally Posted by rai2003
Can u tell me how i can find Foci since i know this is ellipse. Thanks

If you are working on a section about foci of conic sections you should already have access to the relevant formulas. Take a look at this link. (Which I got from a simple web search.) The information you need is under the sections "Focus" and "Polar Form Relative to Center."