How do you find the eccentricity, directrix, focus and classify the conic section $r = \frac{10}{2 - 2 sin θ}$ $r=10/(2-2sintheta)$ ?

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Answer 1 · 2016-12-20T17:31:07

See explanation and graph.

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

$\frac{l}{r} = 1 + e cos (θ + α)$ $l/r=1+e cos (theta+alpha)$

represents

( parabola ellipse hyperbola )

according as

$(e = < > 1)$ $(e =< >1)$

Here, the form is

$\frac{5}{r} = 1 - sin θ = 1 + cos (θ + \frac{π}{2})$ $5/r=1- sin theta =1+cos(theta+pi/2)$ -

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 $θ = α = \frac{π}{2}$ $theta = alpha = pi/2$ .

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

a = 5/2. So, V is $(\frac{5}{2}, - \frac{π}{2})$ $(5/2, -pi/2)$ .

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

from the vertex. So, its equation is

$r sin θ = - 5$ $r sin theta = -5$ ,

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

remembering that focus is at r = 0..

How do you find the eccentricity, directrix, focus and classify the conic section r=102−2sinθr=10/(2-2sintheta)?