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

1 Answer
Dec 20, 2016

Answer:

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..