How do you find the eccentricity, directrix, focus and classify the conic section $r = \frac{8}{4 - 1.6 sin θ}$ $r=8/(4-1.6sintheta)$ ?

Question

8357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T11:30:39

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

$r = \frac{8}{4 - 1.6 sin θ}$ $r= 8/(4-1.6 sin theta)$ , this is similar to standard equation,

$r = \frac{e p}{1 - e sin θ} e, p$ $r= (e p)/(1- e sin theta) e, p$ 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.

$r = \frac{8}{4 - 1.6 sin θ} = \frac{\frac{8}{4}}{\frac{4 - 1.6 sin θ}{4}}$ $r= 8/(4-1.6 sin theta)= (8/4)/((4-1.6 sin theta)/4)$ 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]

How do you find the eccentricity, directrix, focus and classify the conic section r=8/(4-1.6sintheta)r=84−1.6sinθr=8/(4-1.6sintheta)?