Polar representation of conic sections $r(\theta)=\frac1{1 + e \cos\theta}$

Consider a curve given in polar coordinates by $r(\theta) = \dfrac1{1 + e \cos\theta}$, where $e\ge0$.

a) Show that the distance of each point on this curve to the line $x=\frac1e$ is a constant multiple of $r(\theta)$.

b) When $e>1$, show that the curve approaches two asymptotes, find them and sketch the curve. Hint: If the critical angles are $\pm\theta_0$, compute the vertical distance of the point of the curve at angle $\theta = \theta_0+h$ to the line $\theta=\theta_0$, and take a limit using Taylor approximations.

c) Observe that the curve is bounded if and only if $e < 1$. Show that the curve is an ellipse as follows: Let $a$ be the midpoint between the two points intersecting the $x$-axis. Show that $(1-e^2)(x-a)^2 + y^2$ is constant.

d) What happens when $e=1$?

I'm not really sure where to begin with this question. I know that the curve is an ellipse if $e<1$, a parabola if $e=1$ and a hyperbola if $e>1$.

I converted to parametric form, $x=\frac{\cos(\theta)}{1+e\cos(\theta)}$, $y=\frac{\sin(\theta)}{1+e\cos(\theta)}$ but I'm not sure this helps.

I'm also not sure why this is on a differential equation assignment. Any input is greatly appreciated.

EDIT: Okay, so I've got part a) down. Using the Cartesian coordinates, $\left|\frac{1}{e}-\frac{\cos\theta}{1+e\cos\theta}\right|=\left|1+e\cos\theta-e\cos\theta\right|=1$ which is constant.