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

Dec 20, 2016

See explanation and graph.

#### Explanation:

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

$\frac{l}{r} = 1 + e \cos \left(\theta + \alpha\right)$

represents

( parabola ellipse hyperbola )

according as

$\left(e = < > 1\right)$

Here, the form is

$\frac{5}{r} = 1 - \sin \theta = 1 + \cos \left(\theta + \frac{\pi}{2}\right)$-

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 $\theta = \alpha = \frac{\pi}{2}$.

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

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

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 $\left(r , \theta\right)$ upon the axis,

remembering that focus is at r = 0..