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)
- 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
- 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
where is the directrix at a distance to the left of . So, if has polar coordinates and , we have and . Therefore,
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
just as in part (a). For the points on to the right of , we have so implies
(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.