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

Jul 22, 2018

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

#### Explanation:

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

$r = \frac{e p}{1 - e \sin \theta} e , p$ 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.

$r = \frac{8}{4 - 1.6 \sin \theta} = \frac{\frac{8}{4}}{\frac{4 - 1.6 \sin \theta}{4}}$ or

$r = \frac{2}{1 - 0.4 \sin \theta} \therefore e = 0.4 \mathmr{and} e p = 2 \mathmr{and} p = \frac{2}{0.4}$ or

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

pole parallel to the polar axis. Rectangular conversion:

$4 r - 1.6 r \sin \theta = 8 \mathmr{and} 4 \sqrt{{x}^{2} + {y}^{2}} - 1.6 y = 8$.

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

$\left(0 , 0\right)$ and the conic is ellipse .

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