Easy way to draw conics with Bezier control points? Is there an easy way to draw hyperbolas in Adobe Illustrator given semi major axis a and eccentricity e? Presently my method is getting a bunch of x and y coordinates from Microsoft Excel and then assigning these coordinates to anchor points in Adobe Illustrator. This is very laborious and I'm eyeballing how far to pull out the tangent handles.
I use the same labor intensive method to draw parabolas.
I remember 1st derivatives from high school. So I believe I can give the correct slope to my tangent handles. Here's an attempt to draw a parabola:

I know the tangent at the (0, 0) has slope zero and the tangent at (2, 4) has slope 4. But how far do I pull out these tangent lines? By eyeballing I can get the curve to pass through points on the parabola. But eyeballing is imprecise.
Does the 2nd derivative determine how far I'd pull the tangent handles?
If anyone can give me a clue how to draw accurate ellipses and hyperbolas without investing a ton of effort, I'd be grateful.
If anyone's interested here's a few drawings of conic sections I've done in Illustrator.
 A: The steps are as follows:


*

*Draw the tangent lines at the end-points $\mathbf{P}_0$ and $\mathbf{P}_1$ of the hyperbola.

*Draw a line $L$ that's tangent to the hyperbola and parallel to the chord $\mathbf{P}_0\mathbf{P}_1$.

*Intersect the line $L$ with the two tangent lines to get the points $\mathbf{P}_r$ and $\mathbf{P}_s$.

*Compute (or construct) the points
$$
\mathbf{P}_a = \mathbf{P}_0 + \tfrac43(\mathbf{P}_r - \mathbf{P}_0)
$$
$$
\mathbf{P}_b = \mathbf{P}_1 - \tfrac43(\mathbf{P}_1 - \mathbf{P}_s)
$$
In other words, you construct $\mathbf{P}_a$ so that $\mathbf{P}_r$ is three-quarters of the way along $\mathbf{P}_0\mathbf{P}_a$, and similarly for $\mathbf{P}_b$.


The points $\mathbf{P}_a$ and $\mathbf{P}_b$ are the Bézier control points for the desired cubic curve. 
The points $\mathbf{P}_r$ and $\mathbf{P}_s$ are called the Timmer control points of the curve, after their inventor, Henry (Hank) Timmer. The idea was used in internal systems at Douglas Aircraft for many years, and eventually published as "Alternative representation for parametric cubic curves and surfaces." Computer-Aided Design, 12:25-28, 1980. In some ways, the Timmer control points are more intuitive controls than the Bézier points, but they are not well known, and I don't know of any software that uses them today.
This construction will give you a fairly decent approximation of any conic section curve. Both the conic and the cubic will be tangent to the line $L$ at the mid-point $\mathbf{P}_m$ of $\mathbf{P}_r\mathbf{P}_s$. If the conic happens to be a parabola, the cubic will replicate it exactly. The technique works for circular arcs, in particular. Please refer to this question to learn more about the basic technique and a slight improvement that's possible (though probably not worth the trouble in your case).
The picture below illustrates the construction

