Question

How do you find the eccentricity, directrix, focus and classify the conic section `r=10/(2-2sintheta)`?

Answer 1

See explanation and graph.

Explanation:

graph{sqrt(x^2+y^2)-5-y=0 [-10, 10, -5, 5]}

`l/r=1+e cos (theta+alpha)`

represents

( parabola ellipse hyperbola )

according as

`(e =< >1)`

Here, the form is

`5/r=1- sin theta =1+cos(theta+pi/2)`-

The eccentricity e = 1. So, the conis is a parabola.

The semi latus rectum 2a = 5. So, the size of the parabola a = 5/2.

The focus is at the pole r = 0.

The axis of the parabola makes an angle `theta = alpha = pi/2`.

The vertex V is in the opposite direction #theta =- pi/2, at a distance

a = 5/2. So, V is `(5/2, -pi/2)`.

The directrix is perpendicular to the axis at a distance 2a = 5 above

from the vertex. So, its equation is

`r sin theta = -5`,

by projection of the radius to `(r, theta)` upon the axis,

remembering that focus is at r = 0..