I've been trying to figure this out for a bit and haven't found an answer.
the equation is this:
$r=\frac{4}{5-4\sin\theta} $ I know I need to match this up to a conic graph so I divide top and bottom by 5 and get:
$r=\frac{\frac45}{1-\frac45\sin\theta} $
the two vertex will be on the y axis and the directrix (*atleast one directrix) will be 1 since de=$\frac45$ and e=$\frac45$. I keep confusing the formula to find the center/ the foci and the other directrix and I can't seem to find any information online or in my textbook about it. My teacher wants me to graph 2 focus, 2 directrix, 2 vertex, and a center.
also is the way to find these things the same for ellipse and hyperbola?
If we write $r=\sqrt{x^2+y^2}$ and $\sin(\theta)=y/r$ we have:
$$r = \frac{4}{5-4\sin\theta} = \frac{4}{5-y/r}$$
Which can be simplified to:
$$r(5-y/r) = 5r - y = 5 \sqrt{x^2 + y^2} - y = 4$$
Moving things around we find $$25(x^2+y^2) = (4+y)^2 = 16+8y+y^2$$
Finally we have $$25x^2 + 24y^2 - 8y - 16 = 0$$
Complete the square for $y$ to find:
$$25x^2 + 24( y^2 - \frac13 y - \frac23 + \frac{1}{36} - \frac{1}{36}) = 25x^2 + 24(y-\frac16)^2 -16 +\frac{2}{3}$$
Which finally yields:
$$25x^2 + 24(y-1/6)^2 = 16-2/3$$
This is an ellipse, and now you can find what you want by the usual means.