How do you find the eccentricity, directrix, focus and classify the conic section r=8/(4-1.6sintheta)r=841.6sinθ?

1 Answer
Jul 22, 2018

Eccentricity is e=0.4e=0.4 , directrix is y=-5y=5 , focus is at pole (0,0)(0,0) and the conic is ellipse .

Explanation:

r= 8/(4-1.6 sin theta)r=841.6sinθ , this is similar to standard equation,

r= (e p)/(1- e sin theta) e, pr=ep1esinθe,p are eccentricity of conic and

distance of directrix from the focus at pole. (-) sign indicates

that the directrix is below the focus and parallel to the polar axis.

r= 8/(4-1.6 sin theta)= (8/4)/((4-1.6 sin theta)/4)r=841.6sinθ=8441.6sinθ4 or

r=2/(1-0.4 sin theta) :. e= 0.4 and e p=2 or p = 2/0.4 or

p=5 ; e < 1 :. conic is ellipse and directrix is 5 units below the

pole parallel to the polar axis. Rectangular conversion:

4 r -1.6 r sin theta= 8 or 4 sqrt(x^2+y^2)- 1.6 y =8 .

Eccentricity is e=0.4 , directrix is y=-5 , focus is at pole

(0,0) and the conic is ellipse .

graph{4 * sqrt (x^2+y^2) -1.6 y =8 [-10, 10, -5, 5]} [Ans]