Thread: 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
The second question i do not know how to solve it please help me. Thanks

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
The second question i do not know how to solve it please help me. Thanks

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

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

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."

If you really don't have access to these formulas you should read most of this article.

-Dan