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

Question

8114 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T11:30:39

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

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

# r= (e p)/(1- e sin theta) e, 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)# 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]