Answer to Question #114944 in Analytic Geometry for ANJU JAYACHANDRAN

Question

Derive the equation (23) at page 42 of Unit 2, which represents the polar equation of a conic when the directrix L corresponding to a focus F is taken to the right of F.

Answer 1

Let F be a fixed point, the focus, and let L be a fixed line, the directrix, in a plane. A conic section, or conic, is the set of all points P in the plane such that 

$$\frac{PF}{PL}=e,\ e>0$$

where 

e is a fixed positive number, called the eccentricity.

If e=1,

the conic is a parabola.

If e<1, 

the conic is an ellipse.

If e>1, 

the conic is a hyperbola.

By locating a focus at the pole, all conics can be represented by similar equations in the polar coordinate system. In each of these equations,

(r,θ) is a point on the graph of the conic.

e is the eccentricity.

p is the distance between the focus (located at the pole) and the directrix.

For a conic with a focus at the origin, if the directrix is $x=\pm p,$

where 

p is a positive real number, and the eccentricity is a positive real number e,

the conic has a polar equation

$$r=\frac{ e⋅p}{1±cosθ}$$

For a conic with a focus at the origin, if the directrix is $y=\pm p,$

where p is a positive real number, and the eccentricity is a positive real number e,the conic has a polar equation

$$r=\frac{ e⋅p}{1±sinθ}$$

Given that the directrix L

L corresponding to a focus F

F is taken to the right of F.

Directrix $L: x=p$

The distance from the focus to the point P in polar is just r.

PF=r

The distance from the point P to the directrix is  x=p is $PL=p-r\cos{\theta}$ .

 Then

$$e=\frac{PF}{PL}=\frac{r}{p-r\cos{\theta}}$$

Solve for r

$$e.p-e .\cos\theta.r=r$$

$$r=\frac{ e⋅p}{1+cosθ}$$

is the polar equation.