8.5 Polar Equations of Conics

e= 4 ,parabola

e= 1, parabola

e= 1 , ellipse

e =1 ,hyperbola

x = -6 , ellipse

x = 3 , parabola

x = -6 , hyperbola

x = -3 , hyperbola

 $r=\frac{4}{1+\sin\ \theta}$  

 $r=\frac{4}{1-\sin\ \theta}$  

 $r=\frac{4}{1+\cos\ \theta}$  

 $r=\frac{4}{1-\cos\ \theta}$  

$r=\frac{10}{1+\cos\theta}$

$r=\frac{4}{1-\sin\theta}$

$r=\frac{5}{2-\cos\theta}$

$r=\frac{12}{1+3\ \sin\theta}$

$r=\frac{10}{1+\cos\theta}$

$r=\frac{4}{1-\sin\theta}$

$r=\frac{5}{2-\cos\theta}$

$r=\frac{12}{1+3\sin\theta}$

e = 1

e = 3/2

e = 2

e = 1/2

e = 3

 $y^2=16\left(y+4\right)$  

 $x^2=y+4$  

 $x^2=y+64$  

 $x^2=16\left(y+4\right)$  

 $y^2=16\left(x+4\right)$  

 $r=3$  

 $r=\frac{1}{2+\cos\theta}$  

 $r=\frac{1}{4-\sin\theta}$  

 $r=\frac{1}{1+\sin\theta}$  

 $r=1+2\sin\theta$  

$r=\frac{6}{5-2\cos\theta}$

$r=\frac{6}{5+2\cos\theta}$

$r=\frac{5}{5-2\cos\theta}$

$r=\frac{6}{5-2\sin\theta}$

$r=\frac{6}{5+2\sin\theta}$

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

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

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

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