8.5 Polar Equations of Conics

Determine the eccentricity and the type of conic. $r=\frac{20}{4\ +4\ \sin\ \theta}\$ 

• e= 4 ,parabola

• e= 1, parabola

• e= 1 , ellipse

• e =1 ,hyperbola

Tags

CCSS.HSG.GPE.A.1

 $r=\frac{18}{2-6\ \cos\theta}$  
 ​ 
 Determine the equation of the directrix and the type of conic.

• x = -6 , ellipse

• x = 3 , parabola

• x = -6 , hyperbola

• x = -3 , hyperbola

Tags

CCSS.HSG.GPE.A.1

 Write the polar equation of the conic if e = 1 and the directrix is y = 4

•  $r=\frac{4}{1+\sin\ \theta}$  

•  $r=\frac{4}{1-\sin\ \theta}$  

•  $r=\frac{4}{1+\cos\ \theta}$  

•  $r=\frac{4}{1-\cos\ \theta}$  

Tags

CCSS.HSG.GPE.A.1

• $r=\frac{10}{1+\cos\theta}$

• $r=\frac{4}{1-\sin\theta}$

• $r=\frac{5}{2-\cos\theta}$

• $r=\frac{12}{1+3\ \sin\theta}$

• $r=\frac{10}{1+\cos\theta}$

• $r=\frac{4}{1-\sin\theta}$

• $r=\frac{5}{2-\cos\theta}$

• $r=\frac{12}{1+3\sin\theta}$

Determine the eccentricity of the conic.
 $r=\frac{6}{2+3\cos\theta}$  

• e = 1

• e = 3/2

• e = 2

• e = 1/2

• e = 3

Find the cartesian form of the polar equation.
 $r=\frac{8}{1-\sin\theta}$  

•  $y^2=16\left(y+4\right)$  

•  $x^2=y+4$  

•  $x^2=y+64$  

•  $x^2=16\left(y+4\right)$  

•  $y^2=16\left(x+4\right)$  

Which of the following is NOT a polar equation of a conic?

•  $r=3$  

•  $r=\frac{1}{2+\cos\theta}$  

•  $r=\frac{1}{4-\sin\theta}$  

•  $r=\frac{1}{1+\sin\theta}$  

•  $r=1+2\sin\theta$  

Tags

CCSS.HSF-IF.C.7D

Find the polar equation of the conic whose focus is at the pole, eccentricity is 2/5 and the equation of the directrix is

x = -3.

• $r=\frac{6}{5-2\cos\theta}$

• $r=\frac{6}{5+2\cos\theta}$

• $r=\frac{5}{5-2\cos\theta}$

• $r=\frac{6}{5-2\sin\theta}$

• $r=\frac{6}{5+2\sin\theta}$

Tags

CCSS.HSG.GPE.A.2

Write the cartesian equation of the conic represented by the polar equation:
 $r=\frac{20}{4+6\sin\theta}$  

 $Hint:\ Find\ the\ r\ values\ at\ \frac{\pi}{2}and\ \frac{3\pi}{2}.$  

•  $\frac{\left(y-6\right)^2}{9}-\frac{x^2}{5}=1$  

•  $\frac{\left(y+4\right)^2}{9}-\frac{x^2}{5}=1$  
  
  

•  $\frac{\left(y-6\right)^2}{16}-\frac{x^2}{20}=1$  

•  $\frac{\left(y-6\right)^2}{16}-\frac{x^2}{65}=1$