Section 11.7 – Conics in Polar Coordinates

If e < 1, the conic is an ellipse.If e = 1, the conic is a parabola.If e > 1, the conic is a hyperbola.

The ratio of the distance from a fixed point (focus) to a point on the conic to the distance from the point to the directrix is the eccentricity of a conic. It is a constant ratio and is denoted by e. 

Eccentricity

F

P

D

𝑒=𝑃𝐹𝑃𝐷 𝑒 ∙𝑃𝐷=𝑃 𝐹

Section 11.7 – Conics in Polar CoordinatesPolar Equation for a Conic with Eccentricity e

𝑟=𝑘𝑒

1+𝑒𝑐𝑜𝑠𝜃

𝑟=𝑘𝑒

1+𝑒𝑠𝑖𝑛𝜃

The vertical directrix is represented by k.

The horizontal directrix is represented by k.

To use these polar equations, a focus is located at the origin.

Section 11.7 – Conics in Polar CoordinatesGiven the eccentricity and the directrix corresponding to the focus at the origin, find the polar equation.

𝑒=13𝑎𝑛𝑑 𝑦=6

𝑟=𝑘𝑒

1+𝑒𝑠𝑖𝑛𝜃

𝑟=(6 )( 13 )

1+13𝑠𝑖𝑛𝜃

𝑟=(6 )( 13 )

13

(3+𝑠𝑖𝑛𝜃 )

𝑟=6

3+𝑠𝑖𝑛𝜃

Section 11.7 – Conics in Polar CoordinatesGiven the eccentricity and the directrix corresponding to the focus at the origin, find the polar equation.

𝑒=14𝑎𝑛𝑑 𝑥=−2

𝑟=𝑘𝑒

1−𝑒𝑐𝑜𝑠𝜃

𝑟=(2 )( 14 )

1−14𝑐𝑜𝑠𝜃

𝑟=(2 )( 14 )

14

(4−𝑐𝑜𝑠𝜃 )

𝑟=2

4−𝑐𝑜𝑠𝜃

Section 11.7 – Conics in Polar CoordinatesPolar Equation of an Ellipse with Eccentricity e and Major Axis a

𝑘=𝑎𝑒−𝑒𝑎 𝑘=𝑎( 1𝑒 −𝑒)𝑘=𝑎

1𝑒

(1−𝑒2 )

𝑘𝑒=𝑎 (1−𝑒2 )

𝑟=𝑘𝑒

1+𝑒𝑐𝑜𝑠𝜃

𝑟=𝑎 (1−𝑒2 )1+𝑒𝑐𝑜𝑠𝜃

Section 11.7 – Conics in Polar CoordinatesGiven the polar equation, find the directrix that corresponds to the focus at the origin, the polar coordinates of the vertices and the center

𝑒=12

𝑟=6

2+𝑐𝑜𝑠𝜃

𝑟=6

2(1+ 12 𝑐𝑜𝑠 𝜃)𝑟=

3

1+12𝑐𝑜𝑠𝜃

𝑘𝑒=3

𝑘12=3

𝑘=6

3=𝑎(1−( 12 )2)

𝑘𝑒=𝑎 (1−𝑒2 )

3=𝑎( 34 )4=𝑎

Section 11.7 – Conics in Polar Coordinates

y

𝑒=12

𝑎=4𝑎𝑒→412

→8

𝑟=6

2+𝑐𝑜𝑠𝜃

6=6

2+𝑐𝑜𝑠 𝜃

2+𝑐𝑜𝑠 𝜃=1𝑐𝑜𝑠𝜃=−1𝜃=𝜋(6 ,𝜋   )

𝑟=6

2+𝑐𝑜𝑠𝜃

2=6

2+𝑐𝑜𝑠𝜃

2+𝑐𝑜𝑠 𝜃=3𝑐𝑜𝑠𝜃=1𝜃=0(2 ,0   )