Consider a conic section 
with eccentricity 
, a focus at the origin, and with a vertical directrix 
which is a distance 
to the left of 
.
- Prove that if is an ellipse or parabola then every point of lies to the right of and satisfies the equation (in polar coordinates)
![\[ r = \frac{ed}{1 - e \cos \theta}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8e8351d3025c2de5489104ab4fbb91b3_l3.png "Rendered by QuickLaTeX.com")
- Prove that if the conic section is a hyperbola, point on the right branch satisfy the equation in part (a) while points on the left branch satisfy
![\[ r = \frac{-ed}{1 + e \cos \theta}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-732f0f272040b24dd400332641ac7f5d_l3.png "Rendered by QuickLaTeX.com")
- Proof. If is a conic section with eccentricity and focus at the origin, then we know the points on are given by those points
such that
![\[ \lVert X \rVert = e | X \cdot N + d| \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-237c7f67c1cd748eab9067374e7857c1_l3.png "Rendered by QuickLaTeX.com")
where
is the directrix at a distance to the left of . So, if has polar coordinates 
and 
, we have 
and 
. Therefore,![\[ r = e | r \cos \theta + d | \quad \implies \quad r = \frac{ed}{1 - e \cos \theta}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-03e745042e45592f40a3e629c79cdaeb_l3.png "Rendered by QuickLaTeX.com")
if
lies to the right of the directrix (so 
since 
in this case), and since 
, we must have 
- Proof. If is a hyperbola, then we know
and points on the left branch still satisfy
![\[ r = \frac{ed}{1 - e \cos \theta} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ee1270ebab5e21a503448379d0676cd6_l3.png "Rendered by QuickLaTeX.com")
just as in part (a). For the points on
to the right of , we have 
so 
implies![\[ |r \cos \theta + d | = - r \cos \theta - d \quad \implies \quad r = e (-r \cos \theta - d) \quad \implies \quad r = \frac{-ed}{1 + e \cos \theta}. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0fc9c351597854e2cfdda6102935dcce_l3.png "Rendered by QuickLaTeX.com")
(a) We need to prove that every point of C lies to the right of L. Suppose C is an ellipse. Then, we know that 0<e<1. Therefore we have r= e(d +rcosθ) < d+rcosθ . Then, -rcosθ < d-r <d. Thus, since -rcosθ<d, we know that C lies to the right of Directrix.
Similarly, suppose that C is a parabola. Then we know e=1. Therefore we have r= e(d +rcosθ) = d+rcosθ. So -rcosθ = d-r < d. Thus, since -rcosθ<d, we know that C lies to the right of Directrix.